Let $X$ be a general topological space, let $\{x_n\}_{n=1}^\infty\subseteq X$ be a sequence, and let $y\in X$.
Suppose that for every $V\subseteq X$ open neighborhood of $y$, the set $\{n\in\mathbb{N}\mid x_n\in V\}$ is infinite. Does it necessarily follow that there is a subsequence of $\{x_n\}_{n=1}^\infty$ that converges to $y$ ?
(I discovered that the answer is yes under the additional assumption that $X$ is first countable, but I don't know whether this assumption is necessary.)
No: let $\mathfrak U$ be a free ultrafilter on $\mathbb N$. Define a topology on $X = \mathbb N \cup \{\infty\}$, such that each $n \in \mathbb N$ is discrete and the neighborhoods of $\infty$ are the sets $U \cup \{\infty\}$ with $U \in \mathfrak U$. In other words, $X$ is a subspace of $\beta \mathbb N$.
The sequence $(n)_{n \in \mathbb N}$ fulfills the above property for $y := \infty$, but has no convergent subsequence:
Assume $(n)_{n \in N}$ converges to $y$ for infinte $N \subset \mathbb N$. Partition $N$ into $N_1, N_2$, both are infinite. W.l.o.g., $N_1 \notin \mathfrak U$. Then the open neighborhhood $(\mathbb N \setminus N_1) \cup \{\infty\}$ of $\infty$ does not contain almost every element of this subsequence.
Another counterexample that avoids the axiom of choice.
Let $\Sigma$ be the set of sequences $\sigma\colon \mathbb N \to \{0,1\}$ such that $\sigma^{-1}(\{0\})$ has asymptotic density $1$, and let $X=\{0,1\}^\Sigma$, equipped with the product topology ($\{0,1\}$ having the discrete topology). Then for each $n\in\mathbb N$ define $x_n\in X$ to have the value $\sigma(n)$ in each $\sigma$ coordinate, which we denote by $x_n^\sigma=\sigma(n)$.
Let $y\in X$ have the value $0$ in each coordinate, ie. $y^\sigma=0$ for all $\sigma\in \Sigma$.
Now every neighborhood of $y$ contains an open rectangle of the form $$B_{\sigma_1,\cdots,\sigma_m}:=\{x\in X \mid x^{\sigma_1}=\dots=x^{\sigma_m}=0\}\text{,}$$ and such a neighborhood certainly intersects $\{x_n\}$ infinitely often, since $$S:=\bigcap_{k=1}^m\sigma_k^{-1}(\{0\})$$ still has asymptotic density $1$, and for each $n\in S$ we have $x_n \in B_{\sigma_1,\cdots,\sigma_m}$.
On the other hand, given any infinite subset $K\subseteq \mathbb N$, let $\sigma$ be any sequence that is frequently equal to $1$ on $K$, yet still $\sigma^{-1}(\{0\})$ has asymptotic density $1$. (It is trivial to show such a $\sigma$ exists for any infinite subset $K$).
Then $\{x_k^\sigma\mid k\in K\}$ does not converge to $0$, hence $\{x_k\mid k\in K\}$ does not converge to $y$.
