Open subsets are closed under homeomorphism

Question

Definition:

A function $f$ is called a homeomorphism if $f$ is invertible and $f$ and $f^{-1}$ are both continuous.

A subset $S$ of $V$ is called open, If for each $s\in S$, There exists $r \gt 0$ such that $B_r(s)\subseteq S$.

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Question:

If $f:V\to V'$ is a homeomorphism and $S$ is an open subset of $V$, Prove that $f(S)$ is an open subset of $V'$.

Note(My try):
For each $y \in f(S)$, We have: $\exists x \in V\quad f(x)=y$ . So, We want to find $r\gt 0$ such that $B_r(f(x))=\{v \in V:||v-f(x)||<r\}\subseteq f(S)$. Then what? I'm confused... I may be missing something

Answer 1

Let $U$ be open in $V$. Then note that $$ (f^{-1})^{-1}(U)=f(U) $$ is open since $f^{-1}$ is continuous.