Let $M,N$ be manifolds with $\dim M = \dim N$. If $f:M\to N$ is an immersion then $f$ is open.
I thought that I have solved it, but then I thought there could be a mistake:
Let $p\in M$. As $f$ is an immersion, $df_p$ is injective. Hence it is an isomorphism because $\dim M= \dim N$. By the Inverse Function Theorem (for manifolds), $f$ is a local diffeomorphism at $p$.
Let $A\subseteq M$ be an open set. For every $p\in A$, let $U_p\subseteq A$ be an open set such that $f\restriction_{U_p}:U_p\to f(U_p)$ is a diffeomorphism. I thought that as $f\restriction_{U_p}$ is a diffeomorphism, it is a homemorphism, hence $f(U_p)$ is open and $f(A)$ is union of open sets.
But then I remembered that $f(U_p)$ is only open in $f(U_p)$... which we already knew. I mean, $f\restriction_{U_p}$ is open as a function $U_p\to f(U_p)$, where $f(U_p)$ has the subspace topology, so that doesn't mean $f(U_p)$ is open in $N$. Right?