Prove interior point is a topological invariant

Question

Let $(X,\tau)\cong(Y,\tau^*)$, $A\subseteq X$ and $p\in int(A)$. Let $f:(X,\tau)\to (Y,\tau^*)$ be a homeomorphism. To prove that the property of interior point is a topological property we shall show that $f(p)$ belongs to $int(f(A))$. $$p\in int(A)\subseteq A$$ $$f(p)\in f(int(A)) \subseteq f(A)$$ As f is open, $f(int(A))$ is open in $(Y,\tau^*)$ and it is subset of $f(A)$, that is, $f(p)\in f(int(A))\subseteq int(f(A))$ by $int$ definition. Therefore interior point is a topological invariant. Is it my proof right?

Answer 1

If you defined $\operatorname{int}A$ as the largest open subset of $A$ (or if you proved that $\operatorname{int}A$ has that property), then it's totally fine.