First of all, you're missing the assumption that $(X,d)$ is a complete metric space. If we don't assume that, the statement is false, as for $X = \mathbb{Q}$ (usual metric) we have that $(0,1) \cap \mathbb{Q}$ is totally bounded and its closure $[0,1] \cap \mathbb{Q}$ is not compact.
Fact: if $E$ is totally bounded then so is $\overline{E}$.
(Proof: suppose that $r>0$ has been given. Then $E$ is covered by finitely many open balls $B(x_i,\frac{r}{2}), i=1,\ldots n$. Then certainly $E \subseteq \cup_{i=1}^n D(x_i, \frac{r}{2})$, where $D(x,s) = \{y \in X: d(y,x) \le s\}$ is a closed ball. The finite union of closed balls is closed, so $\overline{E} \subseteq \cup_{i=1}^n D(x_i, \frac{r}{2}) \subseteq \cup_{i=1}^n B(x_i,r)$. As $r>0$ was arbitary, $\overline{E}$ is totally bounded.)
So we know that $\overline{E}$ is totally bounded and complete (as a closed subset of the complete $(X,d)$). So $\overline{E}$ is compact (every sequence has a Cauchy subsequence, which converges, and so we have sequential compactness).
Relatively compact always implies totally bounded (a sequence in $E$ has a convergent subsequence with limit in $\overline{E}$ by relative compactness, and a convergent subsequence is a Cauchy subsequence, so $E$ is totally bounded), but for the reverse implication we really need completeness of $(X,d)$, as I showed above.
I think that the argument in the linked post should filter through fine (in fact, it seems constructed for general reflexive spaces).
A reflexive Banach space is separable if and only if its dual is. (The standard result is that $X^*$ separable implies $X$ separable.) Fix a countable dense subsequence $\{\phi_1,\phi_2,\ldots\}\subseteq X^*$, and let $\{a_1,a_2,\ldots\}\subseteq A$ be any sequence. Let them all be bounded by $M$ in norm, since $A$ was bounded. Consider $\{\phi_1(a_1),\phi_2(a_2),\ldots\}\subseteq \mathbb{K}$. This is a bounded sequence in a locally compact space (bounded because all of them have norm less than or equal to $\|\phi_1\|\cdot M$), so it has a convergent subsequence $\phi_1(a_{11}),\phi_2(a_{12}),\ldots$. Now for each $k+1$, similarly extract a convergent subsequence from $\phi_{k+1}(a_{1k}),\phi_{k_1}(a_{2k}),\ldots$. Do a standard diagonal argument and $a_{11},a_{22},\ldots$ is a sequence for which each functional $\phi_i$ converges. Moreover, because the $\phi_i$ were dense, for each $\phi\in X^*$, $\phi(a_{11}),\phi(a_{22}),\ldots$ is convergent. Define the functional $a$ on $X^*$ by $a(\phi)=\lim\limits_{n\to\infty} \phi(a_{nn})$. This is continuous (needs checking) and therefore by reflexivity is represented by some element $x\in X$. Then the sequence $\{a_{nn}\}\xrightarrow{w} x$, and because $A$ was $w$-closed, $x$ is in $A$.
For general spaces, fix your sequence $\{a_1,\ldots\}$ and let $Y$ be its closed linear span. Then $Y$ is separable, and it is a closed subspace of a reflexive space, so itself reflexive (A closed subspace of a reflexive Banach space is reflexive) and the dual is separable. But now we're in the old case, so we extract $x$ with $y^*(a_n)\rightarrow y^*(x)$ for each $y^*\in Y^*$. Now this convergence holds for any linear functional in $X^*$, because the restriction to $Y$ is a linear functional on $Y$.
Best Answer
In any metric space, if $(x_n)_{n\in\mathbb N}$ is a convergent sequence and if $x=\lim_{n\to\infty}x_n$, then $\{x_n\,|\,n\in\mathbb{N}\}$ is relatively compact, since its closure is $\{x_n\,|\,n\in\mathbb{N}\}\cup\{x\}$, which is compact.