= Solution
Let $X=\mathbb R$ with the usual metric and let $Y=\mathbb R$ with the <discrete metric>, and take $f(x)=x$. Every subset of $Y$ is an <open set>, so the image of every open subset of $X$ is open: $f$ is an <open map>.
It is not continuous at any point. For $\epsilon=1/2$, every $\delta>0$ allows a distinct $y$ with $|y-x|<\delta$, but the distance between $f(y)$ and $f(x)$ in the discrete metric is 1. Equivalently, the preimage of the open singleton $\{x\}\subset Y$ is not open in $X$. Thus $\boxed{\text{an open map need not be continuous}}$ even between <metric spaces>.
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