You must have switched things. There is no need for metrizability to prove that open sets are $G_\delta$, nor that closed sets are $F_\sigma$--both of these follow immediately from the definitions of $G_\delta$ and $F_\sigma$.
Now, since a set is $G_\delta$ if and only if its complement is $F_\sigma$, and closed if and only if its complement is open, then we need only prove that every closed set in a metric space is $G_\delta$.
Take any closed set $A$ in a metric space $X$. Let $f(x)=d(x,A)$, so continuity of $f$ means that preimages of open sets are open. In particular, for each integer $n\geq 1$, let $A_n$ be the preimage of $(-1/n,1/n)$. The intersection of all the $A_n$ is then precisely the set of all $x\in X$ such that $f(x)=d(x,A)=0$. Since $A$ is closed, then $d(x,A)=0$ if and only if $x\in A$. Thus, $A=\bigcap_{n=1}^\infty A_n$, so $A$ is $G_\delta$.
Intuitively, if $U$ is open but $f^{-1}(U)$ is not, then $f^{-1}(U)$ contains a point $x_0$ such that for any neighborhood of $x_0$, however small, contains points outside of $f^{-1}(U)$. In other words, one can choose a point $x$ arbitrarily close to $x_0$ such that $f(x)\notin U$, even though $f(x_0)\in U$. For such a point $x$ very, very close to $x_0$, the value of $f(x)$ abruptly “jumps” outside the open set $U$, which is a violation of our intuitive concept of continuity: If $f$ were continuous, then one would expect that for a point $x$ very close to $x_0$, $f(x)$ should be very close to $f(x_0)$.
Formally, I assume we work in metric spaces $(\mathbb X, d_{\mathbb X})$ and $(\mathbb Y,d_{\mathbb Y})$.
The inverse-image definition implies the $\varepsilon$-$\delta$ definition.
Suppose that the inverse-image criterion is satisfied. Let $x_0\in\mathbb X$ and $\varepsilon>0$. Then, the ball $$B_{\mathbb Y}(\varepsilon, f(x_0))\equiv\{y\in\mathbb Y\,|\,d_{\mathbb Y}(y,f(x_0))<\varepsilon\}$$ of radius $\varepsilon$ about $f(x_0)$ is open in $\mathbb Y$, hence $f^{-1}(B_{\mathbb Y}(\varepsilon,f(x_0)))$ is open in $\mathbb X$. Since $x_0\in f^{-1}(B_{\mathbb Y}(\varepsilon,f(x_0)))$, there exists some ball of radius $\delta>0$ about $x_0$ such that $$B_{\mathbb X}(\delta,x_0)\subseteq f^{-1}(B_{\mathbb Y}(\varepsilon,f(x_0)))$$ This is exactly the $\varepsilon$-$\delta$ criterion: if $x\in \mathbb X$ is such that $d_{\mathbb X}(x,x_0)<\delta$, then $d_{\mathbb Y}(f(x),f(x_0))<\varepsilon$.
The $\varepsilon$-$\delta$ definition implies the inverse-image definition.
Suppose that the $\varepsilon$-$\delta$ criterion holds and let $U\subseteq\mathbb Y$ be open. By the definition of openness in metric spaces, there exists for each $y\in U$ some $\varepsilon_y>0$ such that $$B_{\mathbb Y}(\varepsilon_y,y)\subseteq U.$$ In fact, it is not difficult to check that $$U=\bigcup_{y\in U}B_{\mathbb Y}(\varepsilon_y,y).\tag{$\clubsuit$}$$ I now claim that $f^{-1}(U)$ is open in $\mathbb X$. Suppose that $x_0\in f^{-1}(U)$. Then $f(x_0)\in U$, so $f(x_0)\in B_{\mathbb Y}(\varepsilon_{y_0},y_0)$ for some $y_0\in U$ by ($\clubsuit$). [In fact, as @Dominik pointed out in a comment below, one can take $y_0\equiv f(x_0)$. This observation allows to make the derivation that follows a lot simpler.] That is $d_{\mathbb Y}(f(x_0),y_0)<\varepsilon_{y_0}$. Define $$\xi\equiv\varepsilon_{y_0}-d_{\mathbb Y}(f(x_0),y_0)>0.\tag{$\star$}$$ By the $\varepsilon$-$\delta$ definition of continuity, there exists some $\delta>0$ such that $$\text{if }x\in\mathbb X\text{ and }d_{\mathbb X}(x,x_0)<\delta\text{, then }d_{\mathbb Y}(f(x),f(x_0))<\xi.\tag{$\diamondsuit$}$$ I now claim that $$B_{\mathbb X}(\delta,x_0)\subseteq f^{-1}(U),\tag{$\spadesuit$}$$ which will show that $f^{-1}(U)$ is open (since its generic element $x_0$ has a ball around it still in $f^{-1}(U)$), as desired. To this end, let $x\in B_{\mathbb X}(\delta,x_0).$ That is, $d_{\mathbb X}(x,x_0)<\delta$. Then, by ($\diamondsuit$), one has that $$d_{\mathbb Y}(f(x),f(x_0))<\xi.$$ In turn, the triangle inequality and ($\star$) imply that $$d_{\mathbb Y}(f(x),y_0)\leq d_{\mathbb Y}(f(x),f(x_0))+d_{\mathbb Y}(f(x_0),y_0)<\xi+d_{\mathbb Y}(f(x_0),y_0)=\varepsilon_{y_0}.$$ This means that $f(x)\in B_{\mathbb Y}(\varepsilon_{y_0},y_0)\subseteq U$, so that $x\in f^{-1}(U)$. Therefore, ($\spadesuit$) holds, as claimed.
Best Answer
Since the OP's work was reviewed already in the comments, I collect together the entire argument in case future visitors find it useful.
If $f$ is $\varepsilon$-$\delta$-continuous, then it is open-set-continuous. Suppose $f : \mathbb R \to \mathbb R$ is continuous by the $\varepsilon$-$\delta$ definition; we want to prove that it is continuous by the open sets definition.
Take an arbitrary open set $V \subseteq \mathbb R$; we want to prove $f^{-1}(V)$ is open. This is true if $f^{-1}(V)$ is empty, so assume $x \in f^{-1}(V)$. Since $f(x) \in V$ and $V$ is open, there exists some $\varepsilon > 0$ such that $(f(x) - \varepsilon, f(x) + \varepsilon) \subseteq V$. By continuity at $x$, there exists some $\delta > 0$ such that $(x - \delta, x+ \delta) \subseteq f^{-1}(V)$. That is, $x$ is an interior point of $f^{-1}(V)$. Since this is true for arbitrary $x \in f^{-1}(V)$, it follows that $f^{-1}(V)$ is open.
If $f$ is open-set-continuous, then it is $\varepsilon$-$\delta$-continuous. Suppose $f : \mathbb R \to \mathbb R$ is continuous by the open sets definition; we want to prove that it is continuous by the $\varepsilon$-$\delta$ definition.
Fix $x \in \mathbb R$ and $\varepsilon > 0$. Then $(f(x) - \varepsilon, f(x) + \varepsilon)$ is an open set in $\mathbb R$ (containing $f(x)$). By continuity, $U = f^{-1}((f(x) - \varepsilon, f(x) + \varepsilon))$ is an open set in $\mathbb R$. It's easy to see that $U$ contains $x$; then $x$ is an interior point of $U$ by openness of $U$. That is, there exists $\delta >0$ such that $(x - \delta, x+\delta) \subseteq U = f^{-1}((f(x) - \varepsilon, f(x) + \varepsilon))$. Then it follows that $f((x - \delta, x+\delta)) \subseteq (f(x) - \varepsilon, f(x) + \varepsilon)$.