Prove that $f_n$ converges uniformly on $[a,b]$ Let $f_n$ be a sequence of functions defined on $[a,b]$. Suppose that for every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly. Prove that $f_n$ converges uniformly on $[a,b]$
I know that since for every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly, meaning for all $\epsilon >0$ there exist an $N>0$ such that $n>N$ implies $|f_n(c)-f(c)|<\epsilon$.
I know that $c$ is arbitray any where in $[a,b]$, but I don't know how to argue that $f_n$ converges uniformly on $[a,b]$ formally.
ok, here is what I got
Assume that $f_n$ be a sequence of functions defined on $[a,b]$. Suppose that for every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly. Since $[a,b]$ is compact, there exists finitely many number of point $c_1,c_2,c_3, ... \in [a,b]$ such that
$[a,b] \subset \cup _{k=1}^n (c_k - \frac {\delta(c_k)}{2},c_k + \frac {\delta(c_k)}{2})$
since or every $c \in [a,b]$, there exist an interval around $c$ in which $f_n$ converges uniformly, meaning for all $\epsilon >0$ there exist finitely number of  $Ns>0$ such that $n>Ns$ implies $|f_n(c)-f(c)|<\epsilon$. Hence, $f_n$ converges uniformly on $[a,b]$?

You can use a trick similar to mathematical induction but on real line:
Let $I = \{ x \in [a,b]: f_n \mbox{ converges uniformly on } [a,x] \}$
We know:

For every $c \in [a,b]$, there exist an interval around $c$ in which
  $f_n$ converges uniformly

Take $c=a$ in the above statement, and then there exists an interval near $a$ that $f_n$ converges uniformly.
Therefore $I$ is nonempty, $\sup(I)$ exists and let $S = \sup(I)$
We want to prove $S = b$, then we are done.
Suppose $S < b$, 
$f_n$ converges uniformly on $I$ by definition of $I$
Take $c=S$, then there exists an interval $J$ near $S$ that $f_n$ converges uniformly.
Then, take a subset $[S-\varepsilon,S+\varepsilon]$ of $J$ for some $\varepsilon > 0$. 
Then $S+\varepsilon \in I$, contradicting $\sup(I) = S$.
