elementary question about uniform convergence 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?
 A: In the last sentence how do you know that $|x_n-x|<\delta \Rightarrow |f_n(x_n)-f_n(x)|<\frac{\epsilon}{2}?$ Also where did you use uniformly convergence? Note that $x$ is fixed. Your proof is not correct.
For a correct proof use that  $f_n \xrightarrow[n\to\infty]{}f$ uniformly, $f$ is continuous and split $|f_n(x_n)-f(x)|\leq |f_n(x_n)-f(x_n)|+|f(x_n)-f(x)|$ for all $n\geq N$ where $N$  is such that $$|f_n(y)-f(y)|<\dfrac{\epsilon}{2} \text{ and } |f(x_n)-f(x)|<\dfrac{\epsilon}{2}$$ for all $n\geq N$ and $y \in [a,b].$
The assumption of $(f_n)$ being continuous is necessary as the following (counter) example shows.
Consider the sequence of functions $(f_n(x))_{n\in\mathbb N}$ on $[0,1]$ defined  by 
$$f_n(x)=\begin{cases}1 \ \ , \ \ \ x=0\\ \\
\dfrac1n \ \ , \ \ \ x\in \left(0,1\right]{}\end{cases}.$$
The sequence $f_n$ converges uniformly to the function $$f(x)=\begin{cases}1 \ \ , \ \ \ x=0\\ \\
0 \ \ , \ \ \ x\in \left(0,1\right]{}\end{cases}$$
and if $x_n=\dfrac1n$ (or any sequence with $x_n\to0$ and $x_n\neq0 , \ \forall n\in\mathbb N$) then $x_n\xrightarrow[n\to\infty]{}0$ but $f_n(x_n)\not\xrightarrow[n\to\infty]{}f(0).$
