So I've been trying to prove this problem for the past week but I don't know if the path chosen was prudent. I used Baire's Theorem to arrive at my proof for the problem. May someone kindly provide guidance or better proof / correct proof, please?
Let $(X,d)$ be a complete metric space and suppose $A$ and $B$ are both dense and $G_\delta$ subsets of $X$, show $A\cap B\ne \emptyset.$
$\textbf{My Proof:}$ Let $C=A\cap B$. We want to prove that $C$ is dense (and therefore nonempty) since if $C$ is empty and $A$ is dense, then $X=\text{ closure of }A=\emptyset$, a contradiction of $X$ being nonempty. First proving the claim that $C\subseteq X$ is dense if and only if $N\cap C$ is nonempty for every neighborhood $N \subseteq X.$
$(\implies)$ Assuming that $C\subseteq X$ is dense, then $C \cup C'=$closure of $C=X$ for any $x$ in $N$ there's 2 cases:
Case 1): $x$ is in $C$, in which case $x \in N\cap C$ so $N \cap C$ is nonempty
Case 2): $x$ is not in $C$ and $x$ is in $C'$. Then since $x$ is a limit point of $C$ and $N$ is a neighborhood of $x$ there is another point $y$ in $N$ such that $y$ does not equal $x$ and $y$ is in $C$. But then $y\in N \cap C$, so $C \cap N$ is not empty again.
$(\impliedby)$ Assuming that $N \cap C$ is not empty for every neighborhood $N\subseteq X$. We want to show that the closure of $C$ is equal to $X$. Let $x \in X$. If $x$ is in $C$ then we are done. Otherwise, we can assume that $x$ is not in $C$ and we will show that $x$ is a limit point of $C$. For any neighborhood $N$ of $x$ we know that $N \cap C$ is nonempty, so there is some $y$ in the intersection of $N$ and $C$. Since $x$ is not in $C$, and $y$ does not equal $x$, $x$ is a limit point of $C$.
Since this if and only if statement has been proven, now let $N_0$ be any neighborhood in $X$ and we want to show that $N_0 \cap C$ is nonempty. Since $G_1$ is a dense open subset of $X$, we know that $N \cap G_1$ is nonempty, so take any $x_1$ in $N_0 \cap G_1$. Since $N_0 \cap G_1$ is open, there's a neighborhood $N_1$ of $x_1$ and we can shrink $N_1$ if needed to ensure that the diameter of the closure of $N_1$ is less than $1$. Let $E_1=$ the closure of $N_1$. Repeating this process up unitl we define $x_{n-1}, N_{n-1}$, and $E_{n-1}$ are already claimed. Since $G_n$ is a dense open subset of $X$, we have that $N_{n-1} \cap G_n$ is nonempty so take any $x_n$ in $N_{n-1}$ intersect $G_n$. Since $N_{n-1} \cap G_n$ is open, there is a neighborhood $N_n$ of $x_n$ and we can shrink $N_n$ if needed to assure that diameter (closure of $N_n$) is less than or equal to $\frac{1}{n}$. Let $E_n$ be the closure of $N_n$. Then $E_n$ is a collection such that all its intersections equal $\{p\}$ for some point $p$ in $N_0$ and since $E_n \subseteq G_n$ this means there's some point $p$ in each $G_n$ and thus $p\in C=$ intersection of all $G_n$, so $N_0 \cap C$ is nonempty.
This can be made very much shorter. Since $A$ is a $G_\delta$, there is a countable family $\mathscr{U}$ of open sets in $X$ such that $A=\bigcap\mathscr{U}$, and since $A$ is dense in $X$, so clearly is each $U\in\mathscr{U}$. Similarly, there is a countable family $\mathscr{V}$ of dense open subsets of $X$ such that $B=\bigcap\mathscr{V}$. But then
$$A\cap B=\left(\bigcap\mathscr{U}\right)\cap\bigcap\mathscr{V}=\bigcap\left(\mathscr{U}\cup\mathscr{V}\right)$$
is the intersection of a countable family of dense open sets and by the Baire category theorem is dense in $X$ and hence non-empty.
I would define a dense subset of $X$ to be one the meets every non-empty open set, but if you’ve only defined it as one whose closure is $X$, then proving that equivalence is definitely worthwhile, if only for future use.
If you already have the Baire category theorem, there’s no need to reprove it, which is essentially what you’re doing in the last paragraph. Note, by the way, that you never actually said what the sets $G_n$ are.