Note that if $f\colon X\to Y$ is continuous and $A\subseteq Y$ is $G_\delta$, then $f^{-1}(A)$ is also $G_\delta$.
Proof: Let $A=\bigcap U_i$, where $U_i$ are open, and let $V_i=f^{-1}(U_i)$. Since $f$ is continuous $V_i$ is open. Now we will show that $G=\bigcap V_i = f^{-1}(A)$.
Let $x\in G$, then for every $i\in\omega$ we have $x\in V_i$, so $f(x)\in U_i$ for all $i$, and so $f(x)\in A$. Therefore $x\in f^{-1}(A)$.
In the other direction, let $x\in f^{-1}(A)$ then we have $f(x)\in U_i$ for all $i$, therefore $x\in V_i$ for all $i$, so $x\in G$.
The rest of follows from the normality.
Added:
Suppose $X$ is normal and every closed set is $G_\delta$. Let $A$ be a closed set, then $A=\bigcap U_i$ for some $U_i$ open, $i\in\omega$. Without loss of generality $U_i\subseteq U_k$ for $i\ge k$ (otherwise take $U'_i = U_i\cap \bigcup_{j<i} U_j$ as open sets instead).
Since $X$ is normal, we have $f_i\colon X\to[0,1]$ for which $f_i[A]=0$, and $f_i[X\setminus U_i] = 1$.
Now define $f\colon X\to[0,1]$ as:
$$\sum_{k=1}^\infty\frac{f_k(x)}{2^k}$$
This is a continuous function, and $f[A]=0$. Suppose $f(x)=0$ then $f_i(x)=0$ for all $i$, therefore $x\in A$.
Let $F$ and $G$ be disjoint closed sets in a perfectly normal space $X$ as defined in Wikipedia. There is a continuous function $f:X\to [0,1]$ such that $F = f^{-1}[\{0\}]$ and $G = f^{-1}[\{1\}]$. Let $$U = f^{-1}\left[\left[0,\frac13\right)\right]$$ and $$V = f^{-1}\left[\left(\frac23,1\right]\right].$$ Since $f$ is continuous, $U$ and $V$ are open in $X$. Clearly $U \cap V = \varnothing$, $F\subseteq U$, and $G\subseteq V$, so $U$ and $V$ separate the closed sets $F$ and $G$. It follows that $X$ is normal.
Best Answer
In a perfecly normal space $X$ all closed sets are $G_\delta$ and so all open sets are $F_\sigma$.
A space $X$ is completely normal iff all open subspaces are normal.
An $F_\sigma$ subspace of a normal space is normal.
For a proof of 3. see my note here or any decent textbook.
Together these imply what you want.