Proving a map is open I have been trying to solve this "Problem 6." for days and have not made any improvements...

Show that $f:X\to Y$ is open if and only if $f^{-1}[\operatorname{Fr}(B)]\subset\operatorname{Fr}[f^{-1}(B)]$ for each $B\subset Y$.

I have these results proved already:

Let $p:X\to Y$ be an open map. Given any subset $S\subset Y$, and any closed $A$ containing $p^{-1}(S)$, there exists a closed $B\supset S$ such that $p^{-1}(B)\subset A$.
The following four propertis of a map $f:X\to Y$ are equivalent:
  
  
*
  
*$f$ is an open map.
  
*$f[\operatorname{Int}(A)]\subset\operatorname{Int}[f(A)]$ for each $A\subset X$.
  
*$f$ sends each member of a basis for $X$ to an open set in $Y$.
  
*For each $x\in X$ and nbd $U\supset x$, there exists a nbd $W$ in $Y$ such that $f(x)\in W\subset f(U)$.
  

All are from Topology by Dugundji
Thanks for your time!
 A: So you want to show $f: X \to Y$ is an open map iff $$\forall B \subseteq Y: f^{-1}[\partial B] \subseteq \partial f^{-1}[B]\tag{1}$$
where $\partial A$ denotes the boundary of a set $A$ in a space (also denoted by $\operatorname{Bd}(A)$ or $\operatorname{Fr}(A)$ in other texts).
So suppose $f$ is open and $B \subseteq Y$, and let $x \in f^{-1}[\partial B]$ (so $f(x) \in \partial B$), and we want to show that $x \in \partial f^{-1}[B]$, so let $O$ be an open neighbourhood of $x$. We need to show that $O$ intersects both $f^{-1}[B]$ and its complement. The set $f[O]$ is then open (as $f$ is open) and contains $f(x) \in \partial B$ so $f[O]$ intersects $B$ and $Y\setminus B$. This by definition means that for some $x' \in O$, $f(x') \in B$ and for some $x'' \in O$ we have $f(x'') \notin B$, and then $x' \in O \cap f^{-1}[B] \neq \emptyset$ and $x'' \in O \cap X\setminus f^{-1}[B] \neq \emptyset$ and as $O \ni x$ was arbitrary we've shown that $x \in \partial f^{-1}[B]$, and $(1)$ has been shown. Note that the proof really "writes itself": at each stage it's clear what the goal is and where to use the assumption of openness.
Now assume $(1)$ holds, and let $O$ be open in $X$. Note that $B:=f[O]$ is open in  $Y$ iff $\partial B \cap B = \emptyset$ (we're looking for a connection between boundaries and openness to be able to use the assumption $(1)$). So suppose that intersection is not empty, so there is some $x \in O$ such that $f(x) \in f[O]$ and $f(x) \in \partial f[O]$. As $x \in f^{-1}[\partial B]$ by $(1)$ we have that $x \in \partial f^{-1}[B]$ and as $x \in O$, $O$ must intersect $f^{-1}[B]$ and also its complement, but the latter is nonsense, as $O \subseteq f^{-1}[B]$ by definition! So no point in $B \cap \partial B$ exists after all, and so $B$ is open (e.g. because $\partial B^\complement = \partial B \subseteq B^\complement$ so $B^\complement$ is closed).
This shows the equivalence.
