WTS:Let X, Y be metric spaces. If for every open subset U$\subseteq$Y , $F^{-1}$(U) is open in X then F: X$\rightarrow$Y. F is $\textbf{continuous}$ on X .
Proof: Assume for every open subset U in Y, $f^{-1}(U)$ is open in X. Let $\epsilon >0$. Let p $\in X$, and so $f(p) \in Y$. Let $V=N_{\epsilon}(f(p))$ . Then $f^{-1}(V)$ is open, and p$\in f^{-1}(V)$. Hence there exists $\delta>0$ : $N_{\delta}(p)$ $\subset f^{-1}(U)$. Hence we have found delta which depends on epsilon therefore the proof is complete.
The question I have is why can we assume that V= $N_{\epsilon}(f(p))$ moreover, is the proof correct?
Asserting that $f$ is continuous at $p$ means that$$(\forall\varepsilon>0)(\exists\delta>0)(\forall x\in X):d(x,p)<\delta\implies d\bigl(f(x),f(p)\bigr)<\varepsilon.$$Another way of expressing this is: for each $\varepsilon>0$, there is a $\delta>0$ such that$$f^{-1}\bigl(N_\varepsilon\bigl(f(p)\bigr)\supset N_\delta(p).$$That's why we take $V=N_\varepsilon\bigl(f(p)\bigr)$. And, yes, the proof is correct, except that that $U$ should in fact be a $V$.