If SS is an infinite σ\sigma algebra on XX then SS is not countable

I am going over a tutorial in my real analysis course. There is
an proof in which I don’t understand some parts of it.

The proof relates to the following proposition:

($S$ – infinite $\sigma$-algebra on $X$) $\implies$ $S$ is uncountable.

Proof:

Assume: $S=\{A_{i}\}_{i=1}^{+\infty}$. $\forall x\in X: B_{x}:=\cap_{x\in A_{i}}A_{i}$. [Note: $B_{x}\in S$ $\impliedby$ ($B_{x}$ – countable intersection].

Lemma: $B_{x}\cap B_{y}\neq\emptyset \implies B_{x}=B_{y}$.

Proof(of lemma):

$z\in B_{x}\cap B_{y} \implies B_{z}\subseteq B_{x}\cap B_{y}$.

1.$x\not\in B_{z} \implies x\in B_{x}\setminus B_{z} \wedge B_{x}\setminus B_{z} \subset S \wedge B_{x}\setminus B_{z} \subset B_{x}$
(contradiction:$\space$ definition of $B_{x}$) $\implies$ $B_{z}=B_{x}$

2.$y\not\in B_{z} \implies y\in B_{y} \setminus B_{z} \space \wedge \space B_{y}\setminus B_{z} \subset S \space\wedge\space B_{y} \setminus B_{z}\subset B_{y}$(contradiction: definition of $B_{y}$) $\implies$ $B_{z}=B_{y}$ $\implies B_{x}=B_{y} \space \square$

Consider: $\{B_{x}\}_{x\in X}$. If: there are finite sets of the
form $B_{x}$ then: $S$ is a union of a finite number of disjoint sets
$\implies$ $S$ is finite $\implies$ there is an infinite number of sets of
the form $B_{x}$. $\implies$ $|\bigcup\limits_{i\in A \subseteq\mathbb{N}}B_{x_{i}}| \geq \aleph_{0}$.(contradiction) $\square$

There are couple of things I don’t understand in this proof:

1. Why the fact that we found a set ($B_{x}\setminus B_{z}$) in $S$
   containing $x$ and is strictly contained in $B_{x}$ a contradiction
   ?

2. Why if there are only a finite number of different sets of the
   form $B_{x}$ then $S$ is a union of a finite number of disjoint
   sets and is finite ?

Answer

Attribution
Source : Link , Question Author : Belgi , Answer Author : Zev Chonoles