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 $[a,b]$?