 Solution

Suppose continuity at $y$ failed. The negation of the relative-domain epsilon-delta definition gives an ${\epsilon }_0>0$ such that for every $\delta >0$ there is an $x\in [a,b]$ with $|x-y|<\delta$ but $|f(x)-f(y)|\ge {\epsilon }_0$. Choosing $\delta =1/n$ produces a sequence $x_n\to y$ whose image does not converge to $f(y)$, contradicting the hypothesis. Thus $f$ 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.