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{\bf Baire's Theorem. Page 82: 22}
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\begin{lem}
Let $U,V\subset X$ be open and $V$ be dense.  Then $U\cap V$ is open
and dense in $U$.  In particular, $U\cap V\neq\emptyset$.
\end{lem}

\proof
Let $p\in U$ be any given point and $r>0$ any given radius.  Shrinking
$r$ if necessary, we can assume that $N_r(p)\subset U$.  Since $V$
is dense and open, we have that $p\in V'$.  Therefore, there exists
a point $q\in N_r(p)\cap V \subset U\cap V$.  We conclude that 
$U\cap V$ is dense in $U$.
\qed

Now let $G_j, j\in\N$ be the open dense sets given in the problem.  We
will show that $\cap G_j\neq \emptyset$.  By replacing $X$ with
$N_r(p)$ for some $p\in X$ and some radius $r>0$ and replacing
$G_j$ with $G_j\cap N_r(p)$, the above lemma applied to $U=N_r(p)$ and
$V = G_j$ will allow us to conclude that in fact $\cap G_j$ contains a point 
in $N_r(p)$.  That is, $\cap G_j$ is actually dense in $X$.

Working inductively, we construct a sequence of radii $r_n$ converging to zero
and points $p_n$ such that for every $n\in\N$,
\begin{itemize}
\item $\overline{N_{r_n}(p_n)}\subset G_n$;
\item $N_{r_{n+1}}(p_{n+1})\subset N_{r_n}(p_n)$
\end{itemize}
It will then follow from problem 21 that 
$$
\cap G_n \supset \cap \overline{N_{r_n}(p_n)} \neq \emptyset.
$$
So pick $p_1, r_1$ so that $N_{2r_1}(p_1) \subset G_1$.  Clearly,
$\overline{N_{r_1}(p_1)} \subset G_1$.  Given $r_k, p_k$ satisfying
the conditions above, the lemma above allows us to choose $r_{k+1} < r_k/2$ 
and $p_{k+1}$ such that 
$N_{2r_{k+1}}(p_{k+1}) \subset N_{r_k}(p_k) \cap G_{k+1}$.  Then simply
by virtue of our wise choices $N_{r_{k+1}}(p_{k+1})$ satisfies both of the 
above conditions.  The result is a decreasing sequence
$$
\overline{N_{r_1}(p_1)} \supset \overline{N_{r_2}(p_2)} \supset\dots
$$
of closed sets with diameters converging to zero, and the unique point
in the intersection of all these sets is certainly a point in $\cap G_n$.
\qed


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