The theorem 4.8. states that :
A mapping $f$ of s metric space $X$ into a metric space $Y$ is continuous on $X$ if and only if $f^{-1}(V)$ is open in $X$ for every open set $V$ in $Y$
The first part of the proof is straightforward, showing that if $f$ is continuous in $X$ and $V$ is an open set in $Y$ then $f^{-1}(V)$ is open. In the second part we assume that $f^{-1}(V)$ is open in $X$ for every open $V$ in $Y$.
The proof starts in the following way: Fix $p\in X$ and $\epsilon>0$. Let $V$ be the set of all $y\in Y$ such that $d(y, f(p))<\epsilon$.
If $f^{-1}(V)$ is open then there exists $\delta >0$ such that $x\in f^{-1}(V)$ if $d(x,p)<\delta$. But if $x$ belongs to $f^{-1}(V)$ it means that $f(x) \in V$ . By definition it implies $d(f(x), f(p))<\epsilon$.
My doubt is, this seems not very generic, since here $V$ is defined to be an open ball centered at an arbitrary (pre-fixed) point $f(p)$. How to concile it with the "for every open $V$ in $Y$."?
Is there something I am not seeing? Thanks in advance.
EDIT: this holds true for every $p$ in $X$, because every open set contains an open ball of a interior point of the set , that's why we consider just the open ball around $f(p)$?
Rudin is assuming that for every open subset $V$ of $Y$, $f^{-1}(V)$ is an open subset of $X$. Then, he applies this assumption and he is not forced to use it in its general form. It turns out that using it only in the case of open balls is enough.
If he was proving the assertion that for every open subset $V$ of $Y$, $f^{-1}(V)$ is an open subset of $X$, then yes, he would have to prove it in every single case.