 Solution

Fix $x\in X$. By assumption there is a neighbourhood $U$ of $x$ on which $f_n\to f$ with uniform convergence. Each restriction $f_n{|}_U$ is continuous, and a uniform limit of continuous functions is continuous. Thus $f{|}_U$ is continuous, in particular at $x$. Since $x$ was arbitrary, $f$ is continuous on $X$.

Now let $K\subseteq X$ be compact and let $\epsilon >0$. For each $x\in K$, choose a neighbourhood $U_x$ on which convergence is uniform. The sets $U_x$ cover $K$, so compactness gives a finite subcoverFor each $j$, choose $N_j$ such thatTaking $N={\max }_j{N}_j$, every $y\in K$ belongs to one of these finitely many neighbourhoods, and thereforeThus $f_n\to f$ uniformly on every compact subset. This proves the local uniform convergence on compact subsets principle.