Let $f:X\to Y$ be a surjective map such that $\text{int}(f(A))\subset f(\text{int}(A))$ for any $A\in\mathcal{P}(X)$. Show that $f$ is continuous. I'm working on an exercise, stated as follows:

Let $X$ and $Y$ be topological spaces. Let $f:X\to Y$ be a surjective function satisfying the condition that $\text{int}\big(f(A)\big)\subset f\big(\text{int}(A)\big)$ for any subset $A\subseteq X$. Show that $f$ is continuous. [Here $\text{int}(A)$ means the topological interior of $A$, in case there is any confusion.]

I thought I had solved this, but it turns out I was implicitly assuming that $f$ was injective as well. Here is my incorrect "proof":
Let $U\subseteq Y$ be open. Then
$$
U=\text{int}(U)=\text{int}\Big(f\big(f^{-1}(U)\big)\Big)\subset f\Big(\text{int}\big(f^{-1}(U)\big)\Big).
$$
Thus $f^{-1}(U)\subset\text{int}\big(f^{-1}(U)\big)$. (Here is the issue. We only have $f^{-1}(f(A))\supset A$ for every set $A$, with equality when $f$ injective.) Thus $\text{int}\big(f^{-1}(U)\big)=f^{-1}(U)$ and hence $f^{-1}(U)$ is open. Thus $f$ is continuous. 
I've also tried other approaches, but all of them seem to require me to have $f$ injective. I'm starting to think the statement may not even be true. Does anyone have a proof or counterexample of this? Thanks
 A: Let $U \subset Y$ be open. For every $A \subset X$ such that $f\lvert_A$ is a bijection $A \to U$, the condition says
$$f(A) = U = \operatorname{int} f(A) \subset f(\operatorname{int} A),$$
and since $\operatorname{int} A \subset A$, by the injectivity of $f\lvert_A$ it follows that $\operatorname{int} A = A$, i.e. $A$ is open. $f^{-1}(U)$ is the union of such $A$, hence itself open.
A: Let $x \in X$ and $y = f(x)$. Let $V$ be an open neighbourhood of $y$. Now let $U \subseteq X$ be such that $f[U] = V$, $x \in U$ and $f| U $ is injective. (we can pick a point from every fibre from $V$, picking $x$ for $y$). 
Then $y \in V = \operatorname{int}(f[U]) \subseteq f[\operatorname{int}(U)]$.
So $x \in \operatorname{int}(U)$ by 1-1 ness (as $x$ is the only point in $U$ mapping to $y$). So $\operatorname{int}(U)$ is an open neighbourhood of $x$ that maps into $f[U] = V$. So $f$ is continuous at $x$. (Note that we can take $U = f^{-1}[V\setminus \{y\}] \cup \{x\}$, because we only need $x$ to be the only point going to $y$, this makes it more explicit).
This condition is sufficient to prove continuity, but not necessary: look at the projection onto the $x$-axis in the plane which does not have this condition. (take $A$ is a horizontal line e.g. ) 
This is in contrast to $f[\overline{A}] \subseteq \overline{f[A]}$ for all $A$, which characterises continuity..
