Prove the following two statements are equivalent
(a) $f$ is continuous at $x_{0}$
(b) For every open set $V\subset Y$ that contains $f(x_{0})$, there exists an open set $U\subset X$ containing $x_{0}$ such that $f(U)\subset V$.
My solution
Let us prove the implication $(b)\Rightarrow(a)$ first.
For every $\varepsilon > 0$ there is a $\delta > 0$ which correspond to $V = B(f(x_{0}),\varepsilon)$ and $U = B(x_{0},\delta)$. Due to the given hypothesis, \begin{align*} d(x,x_{0}) < \delta \Rightarrow x\in U \Rightarrow f(x)\in f(U)\subset V \Rightarrow d(f(x),f(x_{0})) < \varepsilon \end{align*} and $f$ is continuous at $x_{0}$.
We may now approach the converse implication $(a)\Rightarrow(b)$.
I do not know how to solve this part. Anyone could please help me with this?
Say we have a $V$ is an open containing $f(x_0)$. Then because it is open, we can pick a subset $B(f(x_0),\epsilon) \subseteq V$ for small enough $\epsilon$. But $f$ is continuous so $\exists \delta$ such that $d(x,x_0)<\delta \implies d(f(x),f(x_0)) <\epsilon \implies f(x) \in B(f(x_0),\epsilon) \subseteq V$. Let $U = B(x_0,\delta)$ for that $\delta$ that you found (which depends on $\epsilon$ and $V$). So $f(U) \subseteq B(f(x_0),\epsilon) \subseteq V$.