Is every closed set the set of all subsequential limits of some sequence? Say $S$ is the set of all subsequential limits of $\left\{a_n \right\}$.
I've proved that for a set $A$ in a metric space, there exists $\left\{a_n \right\}$ such that $S=A$ if and only if $A$ is closed.
Is this also valid for general topological spaces?
 A: For completeness, we prove the theorem for separable metric spaces.
First, note that every subspace of a separable metric space is separable.
Let $C$ be a closed subspace of a separable metric space.
Then let $\{p_n\}_{n \in \mathbb{N}}$ be a sequence dense in a closed subspace $C$ (this exists since $C$ is separable).
Let $q$ be the sequence $p_0, p_0, p_1, p_0, p_1, p_2, p_0, p_1, p_2, p_3, ...$.
Claim: every limit of a subsequence of $q$ must be in $C$. Proof: this follows from the fact that every element of $q$ is an element of $C$ and that $C$ is closed.
Claim: every point $x \in C$ is the limit of some subsequence of $q$. Proof: for each $n$, take $i_n$ such that $d(p_{i_n}, x) < 1/n$. Then the sequence $p_{i_1}, p_{i_2}, ..., p_{i_n}, ...$ is a subsequence of $q$ with limit $x$.
So $C$ is exactly the set of subsequential limits of $q$.
Note that the theorem clearly requires the initial space to be separable, since the initial space is itself closed and is thus supposed to be the set of subsequential limits of a sequence, hence separable. In fact, as remarked by Henno, we need $X$ to be hereditarily separable, since every closed subset of $X$ must be separable. Furthermore, we need the space to be sequential for the iff to hold.
We can generalise this theorem to hold in any 2nd-countable sequential $T_2$ space. For any 2nd-countable space is hereditarily separable, since if we have a subspace $C \subseteq X$ and a basis of opens $U_i$, then just take $U_i \cap C$ to be a basis of opens for $C$, so $C$ is 2nd-countable and hence separable.
Let $p$, $q$ be as before. Now for a point $x \in C$, take an enumeration $\{V_i\}_{i \in \mathbb{N}}$ of the portion of the 2nd-countable basis which contains $x$. Then pick $i_n$ such that $p_{i_n} \in \bigcap\limits_{j = 1}^n V_n$ for all $n$. Then $x = \lim\limits_{n \to \infty} p_{i_n}$. And $\{p_{i_n}\}_{n \in \mathbb{N}}$ is a subsequence of $q$.
A: Some thoughts:
First, I think that what you proved is true for separable (or equivalently second countable) metric spaces, otherwise there are simple counterexamples like uncountable discrete spaces.
To extend results like this a larger class of topological spaces, we probably need that $X$ is hereditarily separable (so every closed subset has a countable dense subset) and sequential or maybe even first countable. I think it might be true for sequential hereditarily separable Hausdorff (?) spaces. Sequentiality ensures that sequentially closed and closed are the same in $X$, so that the topology is describable by its convergent sequences.
A: What may be a better way to think about this: If a topological space is "first countable" (i.e., each point has a countable neighborhood basis) then the closure of a set is precisely its set of sequential limit points.
Of course, any metric space is first countable.
A: Note that there are spaces that cannot produce the empty set as the $S$ for some sequence $\{a_n\}$. I assume this is an anomaly and excluding the empty set will make the claim true.
Simply use the interval $[0,1]$ as the underlying set and use as metric the on induced by $\mathbb R$. Any sequence has either a constant infinite subsequence, or an accumulation point, which also provides a converging subsequence.
