I'm trying to show that in a metric space, given a sequence $u_n$, if for any neighbourhood of $x_0$ you can find infinitly number of terms of $u_n$ that are in it, then there exists à subsequence $u_{n_k}$ that converges to $x_0$.
I think that this is an obvious and simple proof, but I can't think of a decisive argument to complete the proof.
This is a rigorous walkthrough of the proof that Shelby hints at in the comments. Perhaps this is the sort of "decisive argument" you are looking for?
Let $X$ denote our metric space. We must find a subsequence $u_{n_k}$ that converges to $x_0$.
For each $k$, consider the open neighborhood $$ B(x_0,1/k) \overset{\text{def}}{=} \lbrace x\in X\,\vert\,d(x_0,x)<1/k\rbrace. $$ By assumption, the set $$ N(k):=\lbrace n\in \mathbb N\,\vert\, u_n\in B(x_0,1/k)\rbrace $$ is infinite. Picking $n_1 := 1$, we define $$ M(k):=N(k)\setminus \lbrace n_j\vert \, j<k\rbrace $$ to avoid repetitions. Then $M(k)$ is nonempty because $N(k)$ is infinite whereas $\lbrace n_j \vert\, j < k\rbrace$ is finite. Since $\mathbb N$ is well-ordered, we may put $$ n_k := \min M(k). $$ Note that by construction $$ u_{n_k}\in B(x_0,1/k). $$ But this says precisely that $$ d(x_0,u_{n_k}) < 1/k. $$ Hence $u_{n_k}\rightarrow x_0$ as $k\rightarrow \infty$.