= Solution
The sequence has <uniform convergence> to $f$ when, for every $\varepsilon>0$, there is an $N$ independent of $x$ such that $|f_n(x)-f(x)|<\varepsilon$ for every $n\ge N$ and every $x\in[a,b]$. Fix $x$ and choose a single $N$ with $\|f_N-f\|_\infty<\varepsilon/3$. By <continuity> of $f_N$, sufficiently small $|y-x|$ within the interval gives $|f_N(y)-f_N(x)|<\varepsilon/3$. The <triangle inequality> then gives
$$
|f(y)-f(x)|\le|f(y)-f_N(y)|+|f_N(y)-f_N(x)|+|f_N(x)-f(x)|<\varepsilon.
$$
This proves <continuity> at every point, including endpoints with the relative topology.
For the moving points, use the just-proved <continuity> of $f$:
$$
\boxed{|f_n(x_n)-f(x)|\le\|f_n-f\|_\infty+|f(x_n)-f(x)|\longrightarrow0.}
$$
This is <uniform convergence at moving evaluation points>: uniform control is what allows the evaluation point to vary with $n$.
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