Is this part of the proof really necessary? Can we immediately say that since $\{a_i\}$ is a cauchy sequence, then it's contained in some closed subset $A_1$ in $M$? Is it necessary to go through mentioning that $\{a_i\}$ is contained by some $B_1$ which may or may not be closed? Sorry if this question is somewhat shallow.

 A: To apply (b) one has to construct a sequence of sets satisfying (b),
so a descending chain of closed sets with diameters going to zero.
The author could streamline this by defining $A_n$ straight away
as the closure of $\{x_m,x_{m+1},x_{m+2},\ldots\}$. By definition
the $A_n$ are closed, and it's almost immediate that $A_n\supseteq A_{n+1}$.
By the Cauchy condition, given 
$\newcommand{\ep}{\varepsilon}\ep>0$ then there is $N$ such that
$d(x_n,x_m)<\ep$ when $m$, $n\ge N$. If $u$, $v\in A_N$ then there
are sequences $(m_k)$ and $(n_k)$ of integers
with each $m_k$, $n_k\ge N$ such that $u=\lim_k x_{m_k}$
and $v=\lim_k x_{n_k}$. Then $d(x_{m_k},x_{n_k})<\ep$
and by the continuity of $d$,  $d(u,v)\le\ep$, so that $A_N$
has diameter $\le\ep$. Then $A_{N+1},A_{N+2},\ldots$ must
also have diameter $\le\ep$. Thus the diameters of the $A_n$
tend to zero, and we can apply (b).
Then (b) gives $y\in\bigcap_k A_k$. Then $x_n$, $y\in A_n$
and by a limiting argument as above, $d(x_n,y)\le\textrm{diam}(A_n)$.
Then $y=\lim_n x_n$.
