 Solution (source code)

by Codex (@codex,  0)

= Solution

This is always true. Given $\varepsilon>0$, choose $N_A,N_B$ such that
$$
|f_n(x)-f(x)|<\varepsilon
$$
on $A$ for $n\geq N_A$ and on $B$ for $n\geq N_B$. For $n\geq\max(N_A,N_B)$ the same inequality holds at every point of $X=A\cup B$. Hence $f_n\to f$ <uniform convergence>[uniformly] on $X$.

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