Let $(M,d)$ and $(M',d')$ be two metric spaces and $f_n:M\mapsto M'$ a sequence of functions and $f:M\mapsto M'$ is a function.
$f_n\rightarrow f$ uniformly $\iff \forall\text{ subsequence } (f_{n_k}),\forall \text{ sequence }(x_k)$ in $M$ $d'(f_{n_k}(x_k),f(x_k))\rightarrow 0$ in $(\Bbb R,|.|)$
Here is how we prove the direction $\big[~f_n$ does not uniformly converge towards $f$ $\big]\implies\exists (n_k)_{k\in\Bbb N}$ and $(x_k)\subset M$ such that $\sup d'(f_{n_k}(x_k),f(x_k))>\delta>0~\forall k\in\Bbb N$ for some $\delta$:
Since $f_n$ does not uniformly converge towards $f$ $$\exists\delta>0 \text{ such that }\forall N\in\Bbb N~\exists n\ge N\text{ with }\sup\limits_{x\in M}d'(f_n(x),f(x))>\delta$$
Then we can recursively construct two sequences $n_k$ and $x_k$ such that $n_{k+1}>n_k$ (or $n_{k+1}\ge n_k+1$) and $d'(f_{n_k}(x_k),f(x_k))>\delta/2$
I get why we can find the appropriate $n_k$ but why shoud the $x_k$ exist? We're working under the assumption that $\forall N~\exists n\ge N$ such that $\sup\limits_{x\in M}d'(f_n(x),f(x))>\delta$.
How I think I understand it is that for each $n_k$ $\sup$ has the property of there being a sequence $(y_l)\subset M$ such that $\lim\limits_{l\rightarrow\infty}d'(f_{n_k}(y_l),f(y_l))>\delta$ and so for $l$ sufficiently large we can find an actual $x_k:=y_l\in M$ such that $d'(f_n(y_l),f(y_l))>\delta/2$.
So for each $n_k$ the $\sup$ is $>\delta$ so there exists $x_k\in M$ such that $d'(f_{n_k}(x_k),f(x_k))>\delta/2$ and the result follows.
Is there any subtlety that I didn't get?