Let $\mathcal{U}$ be an open cover of $(X,d)$. Let $\{d_n: n \in \mathbb{N}\}$ be a dense subset of $X$. For every $n\in \mathbb{N}, q \in \mathbb{Q}$, if there exists some member $U$ of $\mathcal{U}$ that contains $B(d_n, q)$, pick some $U(n,q) \in \mathcal{U}$ that does (there could be plenty of such $U$ so we use AC to pick definite ones). Otherwise $U(n,q) = U_0$ for some fixed $U_0 \in \mathcal{U}$, for definiteness.
Claim: $\{ U(n,q): n \in \mathbb{N}, q \in \mathbb{Q}\}$ is a countable subcover of $\mathcal{U}$.
To see this, let $x \in X$ and find some $U_x \in \mathcal{U}$ that contains $x$ (as we have a cover). Then for some $r>0$ we have $B(x,r) \subset U_x$, and we can also find some $d_{n(x)}$ from the dense subset inside $B(x,\frac{r}{2})$, and next we find a $q(x) \in \mathbb{Q}$ such that $d(x, d_{n(x)}) < q(x) < \frac{r}{2}$.
Note that $x \in B(d_{n(x)}, q(x))$ and $B(d_{n(x)}, q(x)) \subset B(x,r) (\subset U_x)$ by the triangle inequality.
Then $U_x$ witnesses that some member of $\mathcal{U}$ contains $B(d_{n(x)},q)$ and so we know that $x \in B(d_{n(x)}, q(x)) \subset U(n(x), q(x))$ and so $x$ is indeed covered by the subcover, as required.
This proof is quite direct, but it is essentially the same argument one needs to see that all sets $B(d_n, q)$ for $n \in \mathbb{N}, q \in \mathbb{Q}$ form a (countable) base for $X$. So it's not really a simplification per se, but just to show a direct proof is possible.
BTW, it's not too hard to go to hereditarily Lindelöf as well. Essentially the same proof holds: $\mathcal{U}$ is then not necessarily a cover of $X$ and the $x$ is then chosen to be in $\cup \mathcal{U}$ instead of $X$. We use the formulation of hereditarily Lindelöf as : every family of open sets has a countable subfamily with the same union. The proof goes through the same way, really.
No. Consider the Baire space $\omega^\omega$, the product of countably-many copies of $\omega$ with the discrete topology. This space is second-countable and therefore Lindelöf.
Let $\mathcal U_n=\{[t]:t\in\omega^{\leq n}\}$, where $[t]=\{f\in\omega^\omega:f\text{ extends }t\}$. Then given any choices $\mathcal F_n\subseteq\mathcal U_n$ for $n<\omega$, we will find $f\in\omega^\omega$ that's not covered. First pick $f(0)<\omega$ such that $[\langle f(0)\rangle]\not\in\mathcal F_0$. Then if $f(0),\dots,f(n)$ have been chosen such that $[\langle f(0),\dots,f(i)\rangle]\not\in\mathcal F_i$ for each $i\leq n$, we may also choose $f(n+1)<\omega$ such that $[\langle f(0),\dots,f(n),f(n+1)\rangle]\not\in\mathcal F_{n+1}$.
Now consider this $f$. If $f$ extended $t$ for some $[t]\in\mathcal F_n$, this would violate the choice made for $f(n)$. Thus $f\not\in [t]$ for every $[t]\in\mathcal F_n$, so $f$ is not covered. Thus the space is not Menger.
Best Answer
Yes. Let $\mathcal B$ be a countable base for the topology of $X$ and let $\mathcal C$ be any open k-cover of $X$. Let $\mathcal U$ be the collection of all sets $U$ such that (i) $U$ is the union of finitely many elements of $\mathcal B$, and (ii) $U\subseteq C$ for some $C\in\mathcal C$. Then $\mathcal U$ is a countable open k-cover of $X$. For each set $U\in\mathcal U$ choose a set $C_U\in\mathcal C$ with $U\subseteq C_U$; then $\{C_U:U\in\mathcal U\}$ is a countable k-subcover of $\mathcal C$.