Question on Baire's Category Theorem in Rudin Why do we require that $\overline{S}(x_1,r_1) \subset W\cap V_1$? Why is it necessary that the closure is contained here too? Also, formally how do we know the closure can be contained in here?

 A: The idea is this:
Since $V_1$ is dense in $X$, it follows that $V_1$ must have a non empty intersection with any open set in $X$. Take any non empty open set $W$ in $X$. Then $V_1\cap W$ is non empty.
Also, since intersection of two open sets is open, $V_1\cap W$ must be open too. Therefore, there exist some $x_1\in V_1\cap W$ and $r_1'>0$ such that $B(x_1,r_1')\subset V_1\cap W$. Choose $r_1=\frac 12\min (r_1', \frac 13)$ so that $B(x_1,r_1)\subset \color{red}{B[x_1,r_1]}\subset B(x_1,r_1')\subset V_1\cap W $
Here, red colored ball represents closed ball centred at $x_1$ and having radius $r_1$. Since closure of a set is the smallest closed set that contains the set, it follows that $B(x_1,r_1)\subset \overline B (x_1,r_1)\subset B[x_1,r_1]\subset V_1\cap W$.
Now, $V_2$ is open and dense in $X$ so $V_2\cap B(x_1,r_1)$ must be non empty open and dense. As per discussion above, there exist some $x_2\in B(x_1,r_1)\cap V_2$ and $r_2<\frac 1{3^2}$ such that $B(x_2,r_2)\subset \overline B (x_2,r_2)\subset B[x_2,r_2]\subset V_1\cap V_2\cap W$. Note also that $B(x_2,r_2)\subset B(x_1,r_1)$.
Having chosen $x_1,x_2,...,x_{n-1}$ as per the above procedure, choose $x_n\in B(x_{n-1},r_{n-1})$ and $r_n\lt \frac 1{3^n}$ such that $B(x_n,r_n)\subset \overline B(x_n,r_n)\subset V_1\cap V_2\cap ...\cap V_n\cap W$. Note also that $B(x_n,r_n)\subset B(x_{n-1},r_{n-1}),...,B(x_1,r_1)$. Thus, by induction the sequence $(x_n)$ can be constructed.
Clearly, $d(x_{n+1},x_n)<r_n$. Since $r_n<\frac 1{3^n}$, it follows that $d(x_{n+1},x_n)<\frac 1{3^n}$ for all $n$.
For any $m\ge n$, note that $d(x_m,x_n)\le d(x_m,x_{m-1})+d(x_{m-1},d_{m-2})+...+d(x_{n+1},x_n)\leq \sum_{i=m-1}^n\frac 1{3^i}$.
Since the series $\sum \frac 1{3^n}$ converges, RHS can be made arbitrarily small by choosing $m,n$ sufficiently large. It follows that $(x_m)$ is a Cauchy sequence in $X$. Since $X$ is complete, there is some $x\in X$ such that $\lim x_n=x$.
Note that for any $m, x_n\in \overline B(x_m,r_m)$ for every $n>m$. It is known that $x_n$ converges and since $\overline B(x_m,r_m)$ is closed, $x\in \overline B(x_m,r_m)$. Since $m$ is chosen arbitrarily, it follows that $x\in B(x_m,r_m)$ for every $m$. It follows that $x\in V_1\cap V_2\cap ...\cap V_n\cap W$ for every $n$.
That is, $\cap_{n=1}^\infty V_n$ is non empty.
Since $W$ is arbitrary non empty open set, the above also shows that $\cap V_n$ is dense in $X$.
