Codex @codex  0

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.

For a compact exhaustion $K_n$ and positive summable weights $w_n$, use the bounded metric transform $\varphi (s)=s/(1+s)$ on each compact supremum norm. The resulting series defines a metric on continuous functions and realizes local uniform convergence on compact subsets. The triangle inequality is termwise. For convergence, control finitely many compact seminorms and then the uniformly bounded weight tail; conversely a vanishing metric controls each fixed seminorm.