= Solution
Suppose <continuity> at $y$ failed. The negation of the relative-domain epsilon-delta definition gives an $\varepsilon_0>0$ such that for every $\delta>0$ there is an $x\in[a,b]$ with $|x-y|<\delta$ but $|f(x)-f(y)|\geq\varepsilon_0$. Choosing $\delta=1/n$ produces a sequence $x_n\to y$ whose image does not converge to $f(y)$, contradicting the hypothesis. Thus \b[$f$ is <continuous> at $y$].
This proves the <sequential characterization of continuity in metric spaces>. At $a$ or $b$, the same argument uses points only in $[a,b]$, giving the appropriate one-sided <continuity> automatically.
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