Uniform convergence on intervals 
*

*$(f_n)$ converges uniformly on $(a , b)$

*$(f_n)$ converges pointwise at $x = a$ and $x = b$


Prove that $(f_n)$ converges uniformly on $[a , b]$.
I am confused on how to show this, specifically at the endpoints. 
Using Cauchy's criterion and looking at $x = a$, we can write
\begin{align*}
|f_n(a) - f_m(a)| &= |f_n(a) - f_n(x) + f_n(x) - f_m(x) + f_m(x) - f_m(a)| \\
&\leq |f_n(a) - f_n(x)| + |f_n(x) - f_m(x)| + |f_m(x) - f_m(a)|
\end{align*}
but I am lost after that. Maybe that is not the way to go about it.

Given that $(f_n)$ converges uniformly on $(a,b)$ to say $f$, we know that $\forall \epsilon>0$ $\exists \ N_0\in \mathbb{N}$ such that $\forall x\in(a,b)$ $|f_n(x)-f(x)|<\epsilon$.
Also, $(f_n)$ converges pointwise to $f$ at $a$ and $b$, hence $\exists$ $N_1$ and $N_2\in \mathbb{N}$ such that $\forall \epsilon>0$ $|f_n(a)-f(a)|<\epsilon$ and $|f_n(b)-f(b)|<\epsilon$ (respectively).
Choose $N=max\{N_0,N_1,N_2\}$ and see that $(f_n)$ converges uniformly on $[a,b]$.
