Why does $f(U)$ is open for every open $U\subset M$ not imply $f$ is continuous?

Question

Let $f:M \to N$ be a map from a metric space $M$ to a metric space $N$. Does "$f(U)$ is open for every open $U\subset M$" imply $f$ is continuous?

I think it's wrong but I can't find a counter example.

Answer 1

Let $M$ be any non-discrete metric space, for example $\mathbb R$ with the standard metric, and let $N$ be the same set with the discrete metric. If $f:M\to N$ is the identity function, then $f$ is open but not continuous.

Answer 2

Let $f: [0,2] \to [0,1) \cup [2,3]$, \begin{align*} f(x) = \begin{cases} x, & x \in [0,1) \\ x+1, & x \in [1,2] \end{cases} \end{align*}

Now if $U \subset [0,2]$ is open, then $fU$ is open in $[0,1) \cup [2,3]$ (picture helps), but clearly $f$ is not continuous.