The Theorem presented is: Let $(X,d)$ and $(Y,e)$ be metric spaces. A function $f: X \rightarrow Y$ is said to be continuous iff, for each $x \in X$ and each $\epsilon > 0$, there exists $\delta > 0$ such that if $d(x,y) < \delta$, then $e(f(x), f(y)) < \epsilon$.
For the forward implication, I have done: Assume that $f:X \rightarrow Y$ is continuous and let $x \in X$. Let $\epsilon > 0$. Since $B_\epsilon^e(f(x))$ is an open subset of $Y$, $f^{-1}(B_\epsilon^e(f(x)))$ is an open subset of $X$. Because $x \in f^{-1}(B_\epsilon^e(f(x)))$, there exists $\delta > 0$ such that $B_\delta^d(x) \subseteq f^{-1}(B_\epsilon^e(f(x)))$. I'm stuck on the last part, where I need to show that if $d(x,y) < \delta$, then $e(f(x), f(y)) < \epsilon$.
For the other implication: Assume that for each $x \in X$ and each $\epsilon > 0$, there exists a $\delta > 0$ such that if $d(x,y) < \delta$, then $e(f(x), f(y)) < \epsilon$. Let V be an open subset of $Y$. Suppose that $x \in f^{-1}(V)$. Then, $f(x) \in V$. Since $V$ is open, there exists $\epsilon > 0$ for which $(B_\epsilon^e(f(x))) \subseteq V$. By assumption, there exists $\delta > 0$ such that, if $d(x,y) < \delta$, then $e(f(x), f(y)) < \epsilon$. I am stuck showing that $B_\delta^d(x) \subseteq f^{-1}(V)$.
To show that $B_\delta^d(x) \subseteq f^{-1}(V)$, we let $y$ be an arbitrary element in $B_\delta^d(x)$, i.e. $d(x,y) < \delta$. Well, by assumption, $e(f(x),f(y)) < \varepsilon$, so $f(y) \in B_\varepsilon^e(f(x))$. Since $B_\varepsilon^e(f(x)) \subseteq V$, we obtain $f(y) \in V$, i.e. $y \in f^{-1}(V)$.