Using a characterisation of $\min\{\mathfrak{d},\mathfrak{r}\}$ that comes from dualizing a result in Kamburelis' and Węglorz's paper called "Splittings":
A. Kamburelis, B. Węglorz, Splittings, Archive for Mathematical Logic 35, Issue 4 (1996) 263-277, doi:10.1007/s001530050044,
the cardinal invariant $\mathfrak{bidi}$ is equal to that minimum.
To introduce their result, consider the following notion of splitting: given $a\in[\omega]^\omega$ and $\bar{I}=\langle I_n\rangle_{n<\omega}$ an interval partition of $\omega$, $a$ splits $\bar{I}$ iff $a$ contains infinitely many $I_n$'s and also its complement $\omega\smallsetminus a$ contains infinitely many $I_n$'s.
A family $\mathcal{I}$ of interval partitions of $\omega$ is unreapable if no $a\in[\omega]^\omega$ splits all the members of $\mathcal{I}$. On the other hand, say that $A\subseteq[\omega]^\omega$ is interval-splitting if any interval partition of $\omega$ is splitted by some member of $A$.
Kamburelis and Weglorz proved that $\max\{\mathfrak{b},\mathfrak{s}\}$ is equal to the smallest size of an interval-splitting family. On the other hand, the dual proof leads to see that $\min\{\mathfrak{d},\mathfrak{r}\}$ is the smallest size of an unreapable family of interval partitions of $\omega$.
We use this result to obtain $\min\{\mathfrak{d},\mathfrak{r}\}\leq\mathfrak{bidi}$ (the other inequality was pointed in Tsaban's question). Let $Y\subseteq\omega^\omega$ of size $<\min\{\mathfrak{d},\mathfrak{r}\}$, wlog, we may assume that, for any $h\in Y$, $h$ is increasing and $h(0)>0$. Now, consider $h^*\in\omega^\omega$ defined as $h^*(0)=0$ and $h^*(k+1)=h(h^*(k))$ and let $\bar{I}^h$ be the interval partition where $I^h_k=[h^*(2k),h^*(2k+2))$. Therefore (by Kamburelis and Weglorz), there exists an $a\in[\omega]^\omega$ that splits $\bar{I}^h$ for all $h\in Y$. It is very easy to conclude that $Y$ is bi-nondominating (as defined by Tsaban) by $a$, to be more specific, if $I^h_k\subseteq a$ then the $h^*(2k)$-th element of $\omega\smallsetminus a$ is bigger than $h^*(2k+1)=h(h^*(2k))$ and, likewise, if $I^h_k\subseteq \omega\smallsetminus a$ then the $h^*(2k)$-th element of $a$ is bigger than $h(h^*(2k))$.
Naturally, the previous argument would lead to a Tukey embedding that also gives $\max\{\mathfrak{b},\mathfrak{s}\}$ bigger than or equal (actually, just equal) to the dual of $\mathfrak{bidi}$, i.e., the minimal size of a family $A\subseteq[\omega]^\omega$ with the property that, for any $y\in\omega^\omega$ there is an $a\in A$ such that $y$ doesn't dominate both the increasing enumerations of $a$ and $\omega\smallsetminus a$.
A lot of very good observations have already been put into the comments. I'll add one more observation that's too long for a comment:
It is consistent that $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$.
The basic idea is to start with a model of MA + $\neg$CH (where $\mathrm{non}(\mathcal L)$ is already big), and then to force over this model to make $\acute{\mathfrak{m}}$ smaller while leaving $\mathrm{non}(\mathcal L)$ large. The proof uses two facts:
Fact 1: There is a $\sigma$-centered notion of forcing $P$ that does not change the value of $\mathfrak{c}$ and that forces $\acute{\mathfrak{m}} = \aleph_1$.
Fact 2: Suppose $V$ is a model of MA, and $P$ is a $\sigma$-centered notion of forcing (in $V$). Then $P$ does not lower the value of $\mathrm{non}(\mathcal L)$. In particular, if $V \not\models$ CH and if $P$ does not change the value of $\mathfrak{c}$, then $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ in the extension.
My claim follows easily from these two facts. To get a model of $\acute{\mathfrak{m}} < \mathrm{non}(\mathcal L)$, begin with a model of MA + $\mathfrak{c} > \aleph_1$ and force with the notion of forcing $P$ described in Fact 1. The resulting model has $\acute{\mathfrak{m}} = \aleph_1$ (because $P$ makes this true) and $\mathrm{non}(\mathcal L) = \mathfrak{c} > \aleph_1$ (by the "in particular" part of Fact 2). QED
Fact 2 is possibly "well known" but I don't know the standard reference. It was first explained to me by Andreas Blass, who exposits it nicely in the proof of Corollary 49 in this paper.
For Fact 1, the notion of forcing from Theorem 4 in this paper of Arnie Miller does the job. This forcing -- let us call it $P$ -- is $\sigma$-centered and does not change the value of $\mathfrak c$.$^{(*)}$ $P$ is designed to add a partition of $2^\omega$ into $\aleph_1$ closed sets, and it is easy to show (using a genericity argument) that each set in this partition has measure $0$; Miller even points this out in a comment after the proof of his Theorem 4.
There may be a small issue here about partitioning $2^\omega$ instead of $[0,1]$, but we can get around it. Given our partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, first observe that none of them has interior in $2^\omega$ (because then it would fail to have measure $0$). Thus there is a countable, dense $D \subset 2^\omega$ that contains no more than one point of any member of our partition. Recall that if $C$ is a closed measure-zero subset of $2^\omega$, then $C$ minus one point is homeomorphic to a countable disjoint union of Cantor sets, so it can be partitioned into countably many closed measure-zero sets. Thus, we may modify our partition by adding in the sets of the form $\{d\}$ with $d \in D$, and then dividing some other sets into countably many pieces. In this way we obtain a partition of $2^\omega$ into $\aleph_1$ closed measure-zero sets, including all sets of the form $\{d\}$ for $d \in D$. Once we have done this, we observe that there is a measure-preserving homeomorphism from $2^\omega \setminus D$ onto $[0,1] \setminus \mathbb Q$. When we push our partition through this homeomorphism, we obtain a partition of $[0,1]$ into $\aleph_1$ closed measure-zero sets.
$(*)$ Neither of these facts is stated explicitly in the linked paper, but neither is difficult to prove either. (Using Miller's notation from the linked paper:) Each forcing of the form $P(X)$ is $\sigma$-centered, because if two conditions agree on the part that asserts sentences of the form "$[s] \cap C_n = \emptyset$'' then they are compatible (take the union of the other part). It is clear that no $P(X)$ increases $\mathfrak c$, because it is too small ($|P(X)| \leq \mathfrak c$ and it has the c.c.c., so there are only $\mathfrak c$ nice names for reals). Thus $P$, which is a length-$\omega_1$, finite support iteration of forcings of the form $P(X)$, also is $\sigma$-centered and does not increase $\mathfrak c$.
Best Answer
EDIT: In my original post, I showed that $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model. Upon further reflection, I think we can prove a stronger result, with an arguably easier (but completely different) proof:
Theorem: $\mathfrak{x}_{lac} \leq \mathrm{non}(\mathcal N)$.
Note that this implies $\mathfrak{x}_{lac} < \mathrm{cov}(\mathcal N)$ in the random model, and it also implies the consistency of $\mathfrak{x}_{lac} < \mathfrak{d}$.
In addition to this theorem, let me also point out that it is consistent to have $\mathfrak{a} < \mathrm{cov}(\mathcal M)$. (See Corollary 2.6 in this paper of Brendle.) Therefore the lower bound $\mathrm{cov}(\mathcal M) \leq \mathfrak{x}_{lac}$ mentioned in the post already implies $\mathfrak{a}$ is not an upper bound for $\mathfrak{x}_{lac}$.
Proof of the theorem: Suppose we form an infinite set $B$ by choosing from each interval of the form $[2^k,2^{k+1})$ exactly one integer $b_k$ at random, and then taking $B = \{b_k :\, k \in \omega \}$. (By "at random" I mean that we choose with the uniform distribution, so each integer in $[2^k,2^{k+1})$ has probability $1/2^k$ of being selected.) I claim that if $A$ is lacunary, then it is almost surely true that $A \cap B$ is finite.
To see this, fix some $c > 1$ and some $n_0 \in \omega$ such that if $a$ and $b$ are consecutive members of $A$ above $n_0$, then $b/a > c$. If $k$ is large enough that $n_0 < 2^k$, then this implies there are at most $log_c(2)$ members of $[2^k,2^{k+1})$ in $A$. This implies that the probability of choosing $b_k \in A$ is $\log_c(2)/2^k$ when $k$ is large enough. It follows that the probability of there being $>K$ members of $A$ in $B$ (when $K$ is large) is $\sum_{k > K} \log_c(2)/2^k = \log_c(2)/2^K$. Since this goes to $0$ as $K$ goes to infinity, the probability of $A \cap B$ being infinite is $0$.
The idea of choosing $B$ randomly, one point at a time, like this can be formalized by defining a probability measure on a Polish space, where points of the space correspond to possible choices of the sequence of $b_k$'s. What the previous paragraph shows is that in this probability space, the set of all $B$'s with $A \cap B$ infinite form a null set, for any given lacunary set $A$. Hence any non-null subset of this probability space will contain a $B$ with $A \cap B$ finite. Since this holds for any $A$, we see that any non-null subset $X$ of this probability space contains, for any lacunary set $A$, some infinite $B$ with $A \cap B$ finite. The smallest possible size of such a set $X$ is $\mathrm{non}(\mathcal N)$.
QED
One more observation: The strict inequality $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ is also consistent, so this upper bound cannot be improved to an equality. To see this, begin with a model of Martin's Axiom $+ \, \neg \mathsf{CH}$, and then do a legnth-$\omega_1$, finite support iteration of the eventually different reals forcing. It is not difficult to see that this forcing will make $\mathfrak{x}_{lac} = \aleph_1$ in the extension. But the iteration is $\sigma$-centered, and forcing with a $\sigma$-centered poset over a model of $\mathsf{MA}$ does not change the value of $\mathrm{non}(\mathcal N)$. Thus we get $\mathfrak{x}_{lac} < \mathrm{non}(\mathcal N)$ in the extension.
Original post:
It is not provable that $\mathrm{cov}(\mathcal N) \leq \mathfrak{x}_{lac}$, because $\mathrm{cov}(\mathcal N) > \mathfrak{x}_{lac}$ in the random model.
To see this, let me sketch an argument that after forcing to add any number of random reals (in the usual way, via a measure algebra), the collection $[\omega]^\omega \cap V$ of infinite subsets of $\omega$ in the ground model has the property described in the definition of $\mathfrak{x}_{lac}$. That is: for every lacunary set $A \subseteq \omega$ in the extension, there is an infinite $B \subseteq \omega$ in the ground model such that $A \cap B$ is finite.
We work in the ground model. Suppose $\dot A$ is a name for an infinite lacunary set in the extension. There is some fixed $c > 1$ and $n_0 \in \omega$, and some forcing condition $p$, such that $p \Vdash$ if $a$ and $b$ are consecutive elements of $\dot A$ above $n_0$, then $b/a > c$.
I claim that for every $\varepsilon > 0$, there are arbitrarily large $n \in \omega$ such that $m(p \wedge [n \in \dot A]) < \varepsilon$.
(Note: Here I'm using the standard notation for forcing with measure algebras. If $\varphi$ is any well-formed formula in the forcing language, then we write $[\varphi]$ to mean the supremum of all the conditions forcing $\varphi$, and $m([\varphi])$ denotes the measure of this supremum. Roughly, you may think of $m([\varphi])$ as the probability that $\varphi$ ends up being true in the forcing extension.)
To prove my claim, suppose, aiming for a contradiction, that it is false. Then there is some $\varepsilon > 0$ and some $N \in \omega$ such that $m([n \in \dot A]) \geq \varepsilon$ for all $n \geq N$. But this is just another way of saying that the "expected size" of $A \cap \{n\}$ is $\geq\varepsilon$ for all $n \geq N$. By the linearity of expectation, this means the expected size of $A \cap \{k,k+1,\dots,\ell-2,\ell-1\}$ is $\geq (\ell-k)\varepsilon$ whenever $\ell > k > N$. But by our choice of $p$, if $k \geq N$ and $\ell < ck$, then $p \Vdash |A \cap \{k,k+1,\dots,\ell-2,\ell-1\}| \leq 1$. Since $c > 1$, this yields a contradiction for sufficiently large $k$, namely $k > N,2/c\varepsilon$.
Using this claim, we can now find an infinite ground model set almost disjoint from the set named by $\dot A$ in the extension. Using the claim, we may find for each $k \in \omega$ some $n_k > n_{k-1}$ such that $m([n_k \in \dot A]) < m(p)/2^{k+2}$. Now let $p' = p - \bigvee_{k \in \omega}[n_k \in \dot A]$. Then $m(p') \geq m(p) - \sum_{k \in \omega}m([n_k \in \dot A]) > m(p)/2 > 0$, so $p'$ is a condition in our measure algebra, and $p'$ forces $\dot A$ to be disjoint from $\{n_k :\, k \in \omega \}$ (because it extends the complement of each $[n_k \in \dot A]$).
This shows that it is impossible to have a name $\dot A$ for a lacunary set such that $\dot A$ is forced to have infinite intersection with every infinite subset of $\omega$ from the ground model. Therefore there is no such set.