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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.

The function is continuous at $x$ when, for every $\epsilon >0$, there is $\delta >0$ such that $|y-x|<\delta$ implies $|f(y)-f(x)|<\epsilon$.

If this holds and $x_n\to x$, eventually $|x_n-x|<\delta$, so $f(x_n)\to f(x)$. Conversely, if continuity fails, some ${\epsilon }_0>0$ allows, for every $n$, a point $x_n$ with $|x_n-x|<1/n$ but $|f(x_n)-f(x)|\ge {\epsilon }_0$. Then $x_n\to x$ while $f(x_n)\nrightarrow f(x)$, a contradiction. This is the sequential characterization of continuity in metric spaces.

Assume the sequential condition, and let $A\subseteq Y$ be open. If $f^{-1}(A)$ were not open, some $a\in f^{-1}(A)$ would have no ball contained in $f^{-1}(A)$. For each $n$, chooseThen $x_n\to a$, so $f(x_n)\to f(a)\in A$. Because $A$ is open, this forces $f(x_n)\in A$ eventually, a contradiction. Therefore $f^{-1}(A)$ is open. This proves the sequential characterization of continuity in metric spaces.