 Solution

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.