True or False, prove or provide a counterexample.
(a) If B\subseteq X' closed, then $f^{-1}(B)\subseteq X$ is closed.
(b) If $B$ is a bounded subset of X', then $f^{-1}(B)$ is a bounded subset of $X$.
(c) If $A\subseteq X$ and $x_0\in X$ and $f(x_0)\in \operatorname{Acc}(f(A))$ then $x_0\in \operatorname{Acc}(A)$.
My attempts:
(a) True.
Let $B$ be a closed set in X'. Suppose $x_0\in f^{-1}(B)$. Take $\epsilon>0$ : $B_{\epsilon}(f(x_0))\subset B^c$, which is open since $B$ is closed. Then $\exists\delta$ : $f^{-1}(B_{\epsilon}(f(x_0))\subset f^{-1}(B^c)$ So $f^{-1}(B^c)$ is open and $f^{-1}(B^c)$ is closed.
(b) False.
Counterexample: X=X'=\mathbb{R}. Take $f^{-1}(B)=(-\infty,1)$ and $f(x)=\ldots$ not sure if this is actually false.
(c) True.
For $f(x_0)\in \operatorname{Acc}(f(A))$ to be true, $\forall\epsilon >0: B_{\epsilon}(f(x_0))\cap f(A)\neq\emptyset.$ Does that imply that $B_{\epsilon}(x_0)\cap A\neq\emptyset$?