Show that a sequence of functions $\langle f_n \rangle$ fails to converge uniformly to a function $f$ on a set $E$ if and only if there exists some $\varepsilon > 0$ such that a sequence $\langle x_k \rangle$ of points from $E$ and a subsequence $\langle f_{n_k}\rangle$ can be found such that $|f_{n_k}(x_k) - f(x_k)| \geq \varepsilon$.
$(\Leftarrow)$ Trivial?
$(\Rightarrow)$ Suppose $\langle f_n \rangle$ fails to converge uniformly to a function $f$ on $E$. By definition, this means
$$\exists \, \varepsilon >0 , \forall \, N \in \mathbb{N}, |f_n(x) -f(x)| \geq \varepsilon$$
for some $x \in E$ and some $n \geq N$. Choose $\langle f_{n_k} \rangle = \langle f_n \rangle$ and choose $\langle x_k \rangle$ to be any sequence in $E$. Then we are done.