Exercise Show that a sequence of functions $\{f_n\}$ fails to converge uniformly to a function $f$ on a set $E$ if and only if there exists some positive $\varepsilon$ such that a sequence $\{x_k\}$ of points in $E$ and a subsequence $\{f_{n_k}\}$ can be found such that
$$|f_{n_k}(x_k)-f(x_k)| \geq \varepsilon$$
I am struggling to prove the $(\Leftarrow)$ direction.
For the $(\Leftarrow)$ direciton, we suppose that there is $\varepsilon > 0$ such that a sequence $\{x_k\} \subset E$ and $\{f_{n_k}\} \subset \{f_n\}$ such that
$$|f_{n_k}(x_k)-f(x_k)| \geq \varepsilon$$
It is at this point that I get stuck. In the previous direction $\big[(\Rightarrow)\big]$ if you assume that the $f_n$'s are not uniformly convergent on the entire sequence of functions and the entire set $E$, then it follows immediately that any sequence $\{x_k\}$ and subsequence $\{f_{n_k}\}$ also satisfy the conclusion (not uniformly convergent).
But working from the bottom and generalizing outward is tougher. How does one start with a non-uniformly convergent subsequence $\{f_{n_k}\}$ of a sequence of functions $\{f_n\}$ and generalize this to the sequence of functions $\{f_n\}$?
Suppose that, for some $\varepsilon>0$, there is a sequence $(x_n)_{n\in\Bbb N}$ of points of $E$ and a subsequence $(f_{n_k})_{k\in\Bbb N}$ of $(f_n)_{n\in\Bbb N}$ such that $(\forall k\in\Bbb N):\left|f_{n_k}(x_k)-f(x_k)\right|\geqslant\varepsilon$. Then$$(\forall k\in\Bbb N):\sup|f_{n_k}-f|\geqslant\varepsilon.$$Therefore, the sequence $(f_n)_{n\in\Bbb N}$ does not converge uniformly to $f$ because, if it did, $\lim_{n\to\infty}\sup|f_n-f|=0$ and so $\sup|f_n-f|<\varepsilon$ if $n\gg1$.