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The sequence has uniform convergence to $f$ when, for every $\epsilon >0$, there is an $N$ independent of $x$ such that $|f_n(x)-f(x)|<\epsilon$ for every $n\ge N$ and every $x\in [a,b]$. Fix $x$ and choose a single $N$ with $\Vert f_N-f{\Vert }_{\infty }<\epsilon /3$. By continuity of $f_N$, sufficiently small $|y-x|$ within the interval gives $|f_N(y)-f_N(x)|<\epsilon /3$. The triangle inequality then givesThis proves continuity at every point, including endpoints with the relative topology.

For the moving points, use the just-proved continuity of $f$:This is uniform convergence at moving evaluation points: uniform control is what allows the evaluation point to vary with $n$.