 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 $ϵ=1/2$, 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.