Here is exactly what I mean:
Define a function $f:X\rightarrow Y$ from a metric space $X$ to another metric space $Y$. If any subset $A$ of $X$ satisfies $f(\bar A)\subset \overline {f(A)}$, then for any subset $B$ of $Y$, $f^{-1}(\textrm{int } B)\subset \textrm{int }f^{-1}(B)$.
Here, $\bar A$ means the closure of $A$, and $\textrm{int } B$ means the interior of $B$.
I know that, the premise gives rise to the continuity of $f$, then, the conclusion will be easily proved, but I do not like this approach. I want to go straightly from the premise to the conclusion without using the continuity of $f$. Please help me figure out how to do so. Thank you very much!
Assume for reaching a contradiction that there exists a subset $B$ of $Y$ for which $f^{-1}(\text{int }B)\not\subset\text{int }f^{-1}(B)$. Then there exists a $b$ in the interior of $B$ with some $a\in f^{-1}(b)$ such that $a\not\in \text{int } f^{-1}(B)$.
This means that any neighborhood of $a$ contains a point which is mapped outside of $B$ under $f$. This way you can form a sequence $a_n$ converging to $a$ such that $f(a_n)\not \in B$. Now let $A=\{a_n\mid n\in \mathbb{N}\}$ to get the desired contradiction: $$a\in \bar{A}\quad \text{but}\quad b=f(a)\not \in \overline{f(A)}$$ because $b$ is an interior point of $B$ and cannot be approached by the points of $f(A)$ which lie outside of $B$.