if $(x_{n_k})$ is a subsequence of $(x_n)$ which converges to $l$ in a topological space $E$.
How to prove that $l$ is an adherent value of $(x_n)$ ?
I try this:
Let $V$ an open set which contains $l$, the objective is to prove that $\left|\{n\mid x_n \in V\}\right| =+\infty$ .
as $l$ is a limit of $(x_{n_k})$ that is : $$\exists k_0, : \forall k\geq k_0: x_{n_k}\in V$$
I don't know how to continue?
You’re done, aren’t you?
$\{n_k\mid k \ge k_0\}$ is infinite (that's what it means to be a subsequence: it's an injectively indexed cofinal subset of $\mathbb{N}$, hence infinite) and this set is a subset of $\{n \mid x_n \in V\}$, so the latter set is infinite too.
Moreover, if $p$ is a point with a countable local base, say $U_n, n \in \omega$, we can show that if $p$ is an adherent point of $(x_n)$, (so that for all open $O \ni p$ we have $N(O):= \{n: x_n \in O\}$ infinite) then there is a convergent subsequence of $(x_n)$ converging to $p$:
Pick $n_0 = \min(N(U_0))$ which is well-defined. Next pick $n_1 = \min(N(U_0 \cap U_1)\setminus \{0,1,\ldots,n_0\}$, the first index larger than $n_0$ with value in both $U_0$ and $U_1$ (a finite intersection of neighbourhoods of $p$ is still one and so the corresponding $N$-set is infinite, so has points beyond any limit.
Continuing recursively, define $n_{k+1}= \min(N(\bigcap_{i=0}^k U_i) \setminus \{1,\ldots, n_k\}$ and so we get a subsequence $n_k$ (chosen strictly increasingly from the index set) so that for all $m$, $x_{n_m}$ is in all $U_i$ with $i \le m$. It's easy to check that this implies $x_{n_k} \to p$, using that the $U_n$ form a local base at $p$.