Question about details in the Baire Category Theorem. Here is the version of the theorem being used in my class. It is from the textbook Introduction to Topology by Gamelin and Greene.
Beginning of proof
Let $\{U_n \}_{n=1}^{\infty}$ be a sequence of dense open subsets of a complete metric space $X$. Then $\cap_{n=1}^{\infty} E_n$ is also dense in $X$.
We went to show that $\cap_{n=1}^{\infty} U_n$ dense $\iff \overline{\cap_{n=1}^{\infty} U_n}=X$
$\iff \forall x\in X, x\in \overline{\cap_{n=1}^{\infty} U_n}$    ($x$ is adherent to $\cap_{n=1}^{\infty} U_n$)
$\iff \forall x\in X, \forall \epsilon >0$,$\exists y\in B_\epsilon(x)\cap(\cap_{n=1}^{\infty} U_n)$
Proof:
Let $x\in X$, $\epsilon >0$
Since $U_1$ is dense, $x\in X=\overline U_1 \iff \text{x is adherent to}\ U_1$. So we have that $\exists y_1\in B_\epsilon(x)\cap U_1$.
Since $U_1$ is open, $\exists r_1>0$ such that $B_{r_1}(y_1)\subset U_1$
Since $U_2$ is dense, $\exists y_2\in B_{r_1}(y_1) \cap U_2$  ($y_1$ adherent to $U_2$)
Since $U_2$ is open, $\exists r_2$ such that $B_{r_2}(y_2)\subset U_2$
By shrinking $r_1$, we may arrange that $r_1<1$ and $\overline{B_{r_1}(y_1)}\subset U_1\cap B_\epsilon(x)$.
By shrinking $r_2$, we may arrange that $r_2<\frac{1}{2}$ and $\overline{B_{r_2}(y_2)}\subset U_2\cap B_{r_1}(y_1)$.
Continuing in this manner, we get that $\{y_n\}\subset X$ and a sequence $\{r_n\}$ such that $0<r_n<\frac{1}{n}$.
$\overline{B_{r_n}(y_n)}\subset U_n\cap B_{r_{n-1}}(y_{n-1})$. 
It follows that $\overline{B_{r_n}(y_n)}\subset B_{r_{n-1}}(y_{n-1})\subset B_{r_{n-2}}(y_{n-2})\subset ... \subset B_{r_1}(y_1)\subset B_\epsilon(x)$
If $m>n$, $y_m\in B_{r_n}(y_n) \iff d(y_n,y_m)<r_n \to 0$ $\implies \{y_n\}$ is Cauchy.
Since $X$ is complete, $y_n\to y\in X$
Claim: $y\in B_\epsilon (x)\cap(\cap_{n=1}^{\infty} U_n)$ ($\iff y\in \overline{\cap_{n=1}^{\infty}U_n}$)
Proof of claim: If $m>n, y_m\in B_{r_n}(y_n)$ If we remove the first $n$ terms, this sequence is a subset of $B_{r_n}(y_n)$
So $\{y_{n+1},y_{n+2},...\}\to y$
$\implies y\in \overline{B_{r_n}(x_n)}\subset U_n$  (by definition of closure)
$\implies y\in B_\epsilon (x)$
$y\in \cap_{n=1}^{\infty} U_n$ ($y$ is in each $U_n$). 
End of proof
Now, I am confused by the parts
"By shrinking $r_1$, we may arrange that $r_1<1$ and $\overline{B_{r_1}(y_1)}\subset U_1\cap B_\epsilon(x)$.
By shrinking $r_2$, we may arrange that $r_2<\frac{1}{2}$ and $\overline{B_{r_2}(y_2)}\subset U_2\cap B_{r_1}(y_1)$."
Is this because the intersection $U_1\cap B_\epsilon(x)$ is open so we can find $r_1$ small enough to stay in the set? Why can we conclude further that the closure of this open ball is contained in the intersection?
 A: You are right for the first part. For your second question, we may find an $r<1$ such that$B_r(y_2)\subset U_2\cap B_{r_1}(y_1)$. Then we can take $r_2=\frac r2$, since in a metric space, by the triangle inequality, we have $\overline{B_r(x)} \subset B_{2r}(x)$.
A: We have a non-empty open set $B_{r_{n-1}}(y_{n-1}) \cap U_n$ for some $n$ (by the adherence point property, because $U_n$ is dense, so $y_{n-1}$ lies in its closure); the set is open as an intersection of an open ball (so open) and the open $U_n$.
So we can always pick some point $y_n \in B_{r_{n-1}}(y_{n-1}) \cap U_n$. Because that set is open, $y_n$ is an interior point of it, so there is some $s>0$ with 
$$B_s(y_n) \subseteq B_{r_{n-1}}(y_{n-1}) \cap U_n$$
And as $0< s' \le s$ implies $B_{s'}(y_n) \subseteq B_s(y_n)$ (simply by definition: $d(y_n, z) < s'$ then implies $d(y_n, z)< s$ for any $z$) we can take any smaller radius and still have the inclusion of $B_{s'}(y_n)$ into $B_{r_{n-1}}(y_{n-1}) \cap U_n$. So for example take $r_n = \min(\frac{s}{2}, \frac{1}{n+1})$ which is $< s$ (so the inclusion holds) and $\le \frac{1}{n+1} < \frac{1}{n}$ as well.
As $\overline{B_{\frac{s}{2}}(z)} \subseteq B_s(z)$ for any $s>0$ and $z$ in a metric space, and as $r_n \le \frac{s}{2}$, we ensure that 
$$\overline{B_{r_n}(y_n)} \subseteq B_s(y_n) \subseteq B_{r_{n-1}}(y_{n-1}) \cap U_n$$
as required. These steps define $y_n$ and $r_n$ by recursion.
