Proof of closed sets being invariant under inverse images by continuous functions

Proof of closed sets being invariant under inverse images by continuous
functions

I am studying the below portion of the proof, part of which confuses me:
Suppose that $f: A \to \mathbb{R}^m$ is continuous. Also, suppose that $E$
is a closed subset of $\mathbb{R}^m$. Let $B = f^{-1} (E)$. Now, we show
that $\overline{B} = B$. Clearly, $f(B) = f(f^{-1}(E)) \subseteq E$.
Further, that $A$ is closed implies that the closure of any subset C of A
is contained in A $ \\ $
Thus, if $x \in \overline{B}$, then \begin{equation} f(x) \in
f(\overline{B}) \subset \overline{f(B)} \subset \overline{E} = E \nonumber
\end{equation} Thus, $x \in f^{-1} (E) = B$. Therefore, $\overline{B}
\subset B$, and by Theorem 8.32 (i), $B \subset \overline{B}$. Thus $B =
\overline{B}$. Thus, $B$ is closed.
My question is, how do we know that $f(\overline{B}) \subset
\overline{f(B)}$?