Let $a<c<b$, and suppose $\{f_n(x)\}$ is uniformly convergent on $[a,c]$ and on $[c,b]$. Show that it is uniformly convergent on $[a,b]$.
I tried to use the the fact that $d_n=\sup|f_n(x)-f(x)|$ tends to $0$ as $n\to\infty$ when $x$ varies over $[a,c]$, and also when it varies over $[c,b]$.
This gives us that when $x$ is in $[a,b]$, then
$$ \sup|f_n(x)-f(x)| < \max(\sup|f_n(x)-f(x)|: x\in [a,c], \sup|f_n(x)-f(x)|:x
\in [c,b]),$$ which tends to $0$.
The key fact here is that if $u_n \to 0$ and $v_n \to 0$, then $\max(u_n,v_n) \to 0$. This follows from continuity of the function $(x,y) \mapsto \max(x,y)$.
Let $u_n = \sup_{x \in [a,b]} |f_n(x)-f(x)|$, $v_n = \sup_{x \in [b,c]} |f_n(x)-f(x)|$. Since $u_n \to 0, v_n \to 0$, we have $\max(u_n,v_n) \to 0$, and since $\max(u_n,v_n) = \sup_{x \in [a,c]} |f_n(x)-f(x)|$, we have the desired result.
Let $\epsilon >0$. For our first interval $[a,c]$ Pick an $N_1$ such that for all $n\geq N_1$ we have, $$\sup_{x\in [a,c]}|f_n(x)-f(x)|<\epsilon $$
Now pick an $N_2$ such that on the interval $[c,b]$ we have whenever $n\geq N_2$ we have, $$\sup_{x\in [c,b]} |f_n(x)-f(x)|<\epsilon $$
Now pick $N=\max\{N_1,N_2\}$. Then for all $n\geq N$ we have,
$$\sup_{x\in [a,b]} |f_n(x)-f(x)| <\epsilon $$ on the whole interval $[a,b]$.
