= 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 $\varepsilon>0$. 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 subcover
$$
K\subseteq U_{x_1}\cup\cdots\cup U_{x_m}.
$$
For each $j$, choose $N_j$ such that
$$
n\geq N_j,\ y\in U_{x_j}
\quad\Longrightarrow\quad |f_n(y)-f(y)|<\varepsilon.
$$
Taking $N=\max_jN_j$, every $y\in K$ belongs to one of these finitely many neighbourhoods, and therefore
$$
n\geq N\quad\Longrightarrow\quad
\sup_{y\in K}|f_n(y)-f(y)|<\varepsilon.
$$
\b[Thus $f_n\to f$ uniformly on every compact subset. This proves the <local uniform convergence on compact subsets> principle.]
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