Let $(f_n)$ be a sequence of continuous functions on $[a,b]$ that converges uniformly to $f$ on $[a,b]$. Show that if $(x_n) \subseteq[a,b]$ and if $x_n \rightarrow x$, then $\lim f_n(x_n) = f(x) $
My solution: Let $\epsilon > 0$. Take $N \in \mathbb{N}$ such that
$|f_n(x) - f(x)| < \frac{\epsilon}{2}, \forall n>N \text{ and } \forall x \in [a,b].$
Now, since $(f_n)$ is continuous, take $\delta > 0$ such that:
$|x_n - x| < \delta \Rightarrow |f_{N+1}(x_n) - f_{N+1}(x)| < \frac{\epsilon}{2}.$
Now, if $|x_n - x| < \delta$, we must have that:
$|f_n(x_n) - f(x)| \leq |f_n(x_n) - f_n(x)| + |f_n(x) - f(x)| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$
Is this correct?