I believe the answer to this question is yes. I was hoping someone would critique my logic.
Let X be a separable and metrizable space. Then it has a countable dense subset A. Let B be a basis for X.
For every a in A, choose a basis element U in B such that a is in U (if such exists).
This forms a countable collection V. Suppose, for a contradiction, there is an element x that is in no element of V. Then x is in a basis element U. U must contain an element of A since A is dense. But this U is not in our collection V, a contradiction.
Best Answer
For a metric space $(X,d)$ and an infinite cardinal number $\kappa$, the following are equivalent:
$1)\rightarrow 2)$ is obvious, and true for all topological spaces $X$.
$2)\rightarrow 3)$ is true in general as well: Let $\mathcal{N}$ be a network with $\left|\mathcal{N}\right| \le \kappa$. If $\mathcal{U} = \left\{ U_i : i \in I \right\}$ is an open cover of $X$, then for each $x \in X$ we pick $i(x) \in I$ and $N_x \in \mathcal{N}$, such that $x \in N_x \subset U_{i(x)}$. Then $\left\{N_x : x \in X\right\} = \mathcal{N}'$ has cardinality $\le \kappa$, and for each distinct element $A$ from $\mathcal{N}'$ we pick $U(A)$ from $\mathcal{U}$ with $A \subset U(A)$ ($A = N_x$ for some $x$, and we pick $U(A) = U_{i(x)}$). Then $\left\{U(A) : A \in \mathcal{N}'\right\}$ is the required subcover.
$3)\rightarrow 4)$ is always true as well: Let $A$ be closed and discrete. Each $x \in A$ has an open neighbourhood $U_x$ that intersects $A$ in $\{x\}$ only. The open cover $\mathcal{U} = \left\{U_x : x \in A\right\} \cup \{X \setminus A\}$ cannot spare any $U_x$ (or $x$ will not be covered), so the cover $\mathcal{U}$ has cardinality $|A|$ and no subcover of cardinality strictly less than $|A|$. So $|A| \le \kappa$, or we'd have a contradiction with 3).
$4)\rightarrow 5)$ Here we need only perfect normality of $X$, in the sense only that each open set is a countable union of closed sets, or equivalently that each closed set is a $G_\delta$. Let $A$ be discrete, then I claim that $A$ is open in $\overline{A}$.
Proof of claim (needs only that singletons are closed): let $x$ be in $A$ and let $U_x$ be an open neighbourhood of $x$ that intersects $A$ only in $\{x\}$. This $U_x$ has the property that $\overline{A} \cap U_x = \{x\}$ as well: $y \neq x$ and $y \in \overline{A} \cap U_x$, then $U_x\setminus\{x\}$ is an open neighbourhood of $y$, $y \in \overline{A}$ so $U_x\setminus\{x\}$ must intersect $A$, but this can only happen in $\{x\}$, contradiction, so that $\{x\}$ is open in $\overline{A}$.
But then, as $A$ is perfectly normal (being metrisable), $A = \cup_{i \in \mathbb{N}} A_i$ where the $A_i$ are closed in $\overline{A}$ (and thus closed in $X)$. So the $A_i$ are closed and discrete, and by 4) we have $|A_i| \le \kappa$. So $|A| \le \aleph_0 \cdot \kappa = \kappa$, as well.
$5)\rightarrow 6)$ is true for all topological spaces: pick $x_i \in U_i$ for any pairwise disjoint family $\left\{U_i : i \in I\right\}$ of non-empty open sets. By definition we have that $\left\{x_i: i \in I\right\}$ is discrete (as witnessed by the $U_i$), and so $\left|I\right| \le \kappa$, and 6) has been proved.
$6)\rightarrow 7)$ Here we need the metric in a more essential way. For each $n \in \mathbb{N}$, let $D_n$ be a family of points with the property that $x,y \in D_n$ with $x \neq y$ implies $d(x,y) \ge \frac{1}{n}$, and $D_n$ is maximal with that property. Here we use Zorn's lemma, or some equivalent principle. Note that the balls with radius $\frac{1}{2n}$ around the points of $D_n$ are disjoint so that $|D_n| \le \kappa$ by 6).
Let $D = \cup_n D_n$, we claim that $D$ is dense in $X$. We already see that $D$ is of the right size, as $|D| \le \aleph_0 \cdot \kappa = \kappa$. For if $x$ is not in $\overline{D}$, we have that $d(x,\overline{D}) > 0$ and so for some $m \in \mathbb{N}$ we know that $d(x,\overline{D}) > \frac{1}{m}$. But then, for this $m$, $d(x,\overline{D_m}) \ge d(x,\overline{D}) > \frac{1}{m}$ and in particular: $d(x,y) > \frac{1}{m}$ for all $y \in D_m$. But then we could have added $x$ to $D_m$ and would have obtained a strictly larger $D_m$, and this cannot be. So $D$ is dense.
$7)\rightarrow 1)$ This needs the metric "most". Let $D$ be the dense subset of cardinality at most $\kappa$. Let $\mathcal{B} = \left\{B(x,r): x \in D; r \in \mathbb{Q}\right\}$, then $\left|\mathcal{B}\right| \le \aleph_0 \cdot \kappa = \kappa$. I claim that $\mathcal{B}$ is a base for $X$: let $U$ be open and $x \in U$. Some $\epsilon>0 $ exists such that $B(x,e) \subset U$, and as $D$ is dense there is some $y \in D$ in $B(x,\frac{\epsilon}{3})$. Now pick $r \in \mathbb{Q}$ such that $\frac{\epsilon}{3} < r < \frac{\epsilon}{2}$, then $x \in B(y,r)$ (which is from $\mathcal{B}$) and $B(y,r) \subset B(x,\epsilon)$: if for some $z$, $d(z,y) < r$ then $d(z,x) \le d(z,y) + d(y,x) < r + r < \epsilon$, and so there is a $B_x = B(y,r)$ from $\mathcal{B}$ such that $x \in B_x \subset U$, as required for a base.
This concludes the proof of the equivalence, which shows that weight, network weight, Lindelöf number, extent, cellularity and other cardinal invariants are all the same for metrisable spaces. In particular, their countable variants coincide, so separability is equivalent to second countability, etc.