Definition: $A \subseteq X$ for a metric space $X$ has the Baire Property if for any sequence of sets {$V_{n}$} for $n \geq 1$ that are dense and open in $A$, $$cl(\cap V_{n}) \cap A = A $$ for all $n \geq 1$. That is, $\cap V_{n}$ is dense in $A$.
Question: Now suppose $G_{1}$ and $G_{2}$ are two open, dense sets of metric space $X$. Prove that $ G_{1} \cap G_{2}$ has the Baire property.
To prove this, I need to show that for any sequence, say {$V_{n}$} s.t. each $V_{n}$ is open and dense in $ G_{1} \cap G_{2}$,
$cl(\cap V_{n}) \cap (G_{1} \cap G_{2}) = (G_{1} \cap G_{2}) $
Attempt $V_{n}$ dense in $G_{1} \cap G_{2}$. So $cl(V_{n})\cap (G_{1} \cap G_{2}) = (G_{1} \cap G_{2}) $. Which means $(G_{1} \cap G_{2}) \subseteq cl(V_{n}) $. Thus, $cl(G_{1} \cap G_{2}) \subseteq cl(cl(V_{n})) = cl(V_{n})$.
And since $(G_{1} \cap G_{2}) \subseteq cl(G_{1} \cap G_{2})$, we have $(G_{1} \cap G_{2}) \subseteq cl(V_{n})$.
How do I proceed from here?
You just need to show that if $X$ is Baire, so is any open dense subspace $O$ of $X$. This is straightforward from the definition.
Then note that a finite intersection of dense open sets is still dense and open.
Note that in your statement $X$ needs to be Baire itself, or we could trivially take $X= G_1 = G_2 = \mathbb{Q}$ in the usual topology and have a counterexample.