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 $f(\overline{B}) \subset \overline{f(B)}$?
 A: Suppose that $x\in\operatorname{cl}B$; then there is a sequence $\langle x_n:n\in\Bbb N\rangle$ in $B$ that converges to $x$. The function $f$ is continuous, so the sequence $\langle f(x_n):n\in\Bbb N\rangle$ converges to $f(x)$. Clearly $f(x_n)\in f[B]$ for each $n\in\Bbb N$, so $\langle f(x_n):n\in\Bbb N\rangle$ is a sequence in $f[B]$ converging to $f(x)$, and therefore $f(x)\in\operatorname{cl}f[B]$. Since $x$ was an arbitrary point of $\operatorname{cl}B$, it follows that $f[\operatorname{cl}B]\subseteq\operatorname{cl}f[B]$.
Added: The foregoing argument works only if $A$ is first countable, as is the case, for instance, if it’s a metric space; I original (mis)read the question as saying that $A$ was a subset of a Euclidean (hence metric) space. In the general case one has to work a little harder, but the result remains true.
Again suppose that $x\in\operatorname{cl}B$, and suppose that $f(x)\notin\operatorname{cl}f[B]$; then there is an $\epsilon>0$ such that $N(f(x),\epsilon)\cap f[B]=\varnothing$, where $N(y,r)$ is the open ball in $\Bbb R^n$ of radius $r$ and centred at $y$. Since $f$ is continuous, there is an open nbhd $U$ of $x$ in $A$ such that $f[U]\subseteq N(f(x),\epsilon)$, and therefore $f[U]\cap f[B]=\varnothing$. But that clearly implies that $U\cap B=\varnothing$, contradicting the assumption that $x\in\operatorname{cl}B$. Thus, we must have $f(x)\in\operatorname{cl}f[B]$ and hence $f[\operatorname{cl}B]\subseteq\operatorname{cl}f[B]$.
