Product of Baire spaces Hi¡ I have troubles with the next exercise. I'm so stuck. Any hint? 

Let $X$ and $Y$ be a Baire spaces. Prove that if $X\times Y$ is of second category in itself, then, $X\times Y$ is a Baire space.

I have some ideas, but, really I don't know how can I conclude. First, I thought in the definition. I take $\{ U_n\}_{n\in\mathbb{N}}$ a family of open and dense sets in $X\times Y$. Because $X\times Y$ is of second category in itself, we can conclude that $\displaystyle\bigcap_{n\in\mathbb{N}}U_n\neq\emptyset$. We need to prove now that $\displaystyle\bigcap_{n\in\mathbb{N}}U_n$ is a dense set in $X\times Y$. By contradiction, if we take a non empty open set $V$ of $X\times U$ such that $V\cap\left( \displaystyle\bigcap_{n\in\mathbb{N}}U_n\right)=\emptyset$, then, take a basic open $A\times B\subseteq V$. In this way, $(A\times B)\cap\left( \displaystyle\bigcap_{n\in\mathbb{N}}U_n\right)=\emptyset$. I think that $\Pi_{X}\left[ \displaystyle\bigcap_{n\in\mathbb{N}}U_n\right]$ (projection over $X$) is a dense set in $X$ and $\Pi_{Y}\left[ \displaystyle\bigcap_{n\in\mathbb{N}}U_n\right]$ (projection over $Y$) is a dense set in $Y$. In this way we obtain the desired contradiction because both pojections intersects $A$ and $B$ and then, $\displaystyle\bigcap_{n\in\mathbb{N}}U_n$ intersects $V$. But I don't know how can I prove the density of the projections (is it true?)
The other way is prove that every open set of $X\times Y$ is of second category in $X\times Y$. Again, by contradiction. Suposse that there exist $V$ open set in $X\times Y$ such that $V$ is of firts category in $X\times Y$, then, $V=\displaystyle\bigcup_{n\in\mathbb{N}}C_n$ with every $C_n$ a nowhere dense set. I think then that the projection of $V$ over $X$ is again a first category set, but this will be a contradiction (clearly, if it is true). 
Any hint? I really appreciate any help you can provide me. 
 A: The answer is thanks to the user bof:
Suppose $X_0$ and $Y_0$ are Baire spaces such that $X_0\times Y_0$ is not a Baire space. Such spaces there exist by this, Theorem 2.3, page 121. Let $X=X_0+\mathbb{R}$ (topological sum) and let $Y=Y_0+\mathbb{R}$. $X$ is a Baire space. Suposse $U_n$ is open and dense in $X$. Then $U_n\cap X_0$ is open and dense in $X_0$, and $U_n\cap \mathbb{R}$ is open and dense in $\mathbb{R}$. Since $X_0$ and $\mathbb{R}$ are Baire spaces, $\displaystyle\bigcap_{n\in\mathbb{N}}\left(U_n\cap X_0 \right)$ is dense in $X_0$ and $\displaystyle\bigcap_{n\in\mathbb{N}}\left( U_n\cap\mathbb{R}\right)$ is dense in $\mathbb{R}$. If follows that $\displaystyle\bigcap_{n\in\mathbb{N}}U_n=\left(\displaystyle\bigcap_{n\in\mathbb{N}}\left(U_n\cap X_0 \right)\right)\cup\left( \displaystyle\bigcap_{n\in\mathbb{N}}\left( U_n\cap\mathbb{R}\right)\right)$ is dense in $X$. 
Now, $X\times Y$ is of second category in itself. Suposse $X\times Y$ is of the first category in itself. Then, $X\times Y=\displaystyle\bigcap_{n\in\mathbb{N}}F_n$ where $F_N$ is a nowhere dense subset of $X\times Y$. Since $\mathbb{R}\times\mathbb{R}$ is open in $X\times Y$, it follows that $F_n\cap\left(\mathbb{R}\times\mathbb{R}\right)$ is nowhere dense in $\mathbb{R}\times\mathbb{R}$. But this means that $\mathbb{R}\times\mathbb{R}$ is of the first category in itself, contradicting the Baire category theorem.
Finally, I assumed that $X_0\times Y_0$ was not a Baire space. Now, $X_0\times Y_0$ is open in $X\times Y$, and an open subspace of a Baire space is a Baire space, so $X\times Y$ can't be a Baire space.
