continuous $f:X\rightarrow X'$, $X,X'$ metric spaces properties 
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  $B_{\epsilon}(x_0)\cap A\neq\emptyset$?
 A: (a) is indeed true, but your proof as it stands will not work, because you picked $x_0 \in f^{-1}(B)$, so $f(x_0)$ will be in $B$, not $B^c$. You want to show $f^{-1}(B)^c$ is open, I think, so you need to pick $x_0$ in that set. Then indeed $f(x_0) \in B^c$, so as this set is open there is some $\epsilon > 0$ such that $B_{\epsilon}(f(x_0) \subset B^c$, as you then claim. Then you do not use the actual $\delta > 0$, but you probably mean to apply continuity of $f$ at $x_0$, so there is a $\delta > 0$ such that $f( B_{\delta}(x_0) ) \subset B_{\epsilon}(f(x_0))$. Then you skip a minor step: note that we now know that for every $x' \in B_{\delta}(x_0)$, $f(x')$ is in $B_{\epsilon}(f(x_0)) \subset B^c$, so that indeed $x' \notin f^{-1}(B)$, and this shows that indeed, $B_{\delta}(x_0) \subset f^{-1}(B)^c$, showing that the latter set is open, because every point in it is an interior point.
For (b) consider a constant map from say $\mathbb{R}$ to itself. The image is a singleton $B = \{c\}$, and the inverse image of $B$ is all of $\mathbb{R}$, unbounded.
For (c), the opposite is true: $x_0 \in \mbox{Acc}(A)$ implies $f(x_0) \in \mbox{Acc}(f(A))$, and this property characterizes continuity for $f$ (when this holds for all subsets $A$ and all points $x_0$). As it stands it fails: try the function $f(x) = x$ from $\mathbb{R}$ in the discrete metric ($d(x,y) = 1$ for $x \neq y$, $0$ otherwise) to $\mathbb{R}$ in the standard metric. Then $A = (0,1)$ has only $A$ as its accumulation points, as all sets are closed in the discrete metric, but in the image it has $[0,1]$ as its accumulation points, so taking $x_0 = 0$ or $1$ will be a counterexample.
In fact, this property characterizes closed maps, functions that map closed sets to closed sets.
