Let $(0, +\infty)$ and $[-1, 1]$ have the subspace topologies these sets inherit from the set $\mathbb{R}$ of real numbers with the usual topology.
I know that the function $f \colon (0, +\infty) \longrightarrow [-1, 1]$ defined by $$ f(x) = \sin \frac1x $$ is continuous but not open.
How to explicitly show that this map is not a closed map either?
Let: $$A = \left\{\frac{1}{(4n+1)\frac{\pi}{2} + \frac{1}{n}} : n \in \Bbb{N}\right\}.$$ This set is closed in $(0, \infty)$, as it is the image of a sequence convergent in the larger space $\Bbb{R}$, but not in $(0, \infty)$. We then have \begin{align} f(A) &= \left\{\sin\left((4n+1)\frac{\pi}{2} + \frac{1}{n}\right) : n \in \Bbb{N}\right\} \\ &= \left\{\sin\left(\frac{\pi}{2} + \frac{1}{n}\right) : n \in \Bbb{N}\right\}, \end{align} the closure of which contains $1$. However, the set itself does not contain $1$ (as $\pi$ is irrational). Thus, $f(A)$ is not closed, so $f$ is not a closed map.