If S is an infinite σ algebra on X then S is not countable

Suppose $S$ is countable. Inductively, we may construct a collection of disjoint subsets $T$ in $S$ as follow:  


Step 1. Pick a non-empty set $A_1 \in S$. This is legitimate since $S$ is infinite. Define $T:={A_1}$
Step n+1. Let $n\in\mathbb{N}$. The collection of sets 

\[S-T=S-\{A_i: i=1, 2, ..., n\}\]

is infinite. Thus we may pick a non-empty set $B_{n+1}\in S-T$. Set $A_{n+1}=B_{n+1}-\bigcup_{i=1}^n A_i$. Define $T:=T\cup\{A_{n+1}\}$.

It is clear that the sets in $T$ are disjoint. We want to use $T$ to give uncountable sets in $S$. There is a unique binary representation the number in $[0, 1]$.  We adjust this representation so that $1=0.\bar{1}$. Define a function $f: [0, 1] \rightarrow S$ by 
\[f(0.a_1a_2a_3...)=\bigcup_{A_i\in T'} A_i\]
, where $A_i \in T$ iff $a_i =1$. 
Since the representation is unique, this function is unique. Let $r,s\in[0, 1]$ Write $r=0.r_1r_2...$ and $s=0.s_1s_2...$. If $r \neq s$, then the $r_i\neq s_i$ for some $i\in\mathbb{N}$. 
If $r_i=1$ and $s_i=0$, then $A_i\subset f(r)-f(s)$, meaning $f(r)\neq f(s)$. If $r_i=0$ and $s_i=1$, we just need to change the row of $r$ and $s$. This gives a injective map from $[0, 1]$ to $S$. Because $[0, 1]$ is uncountable, $S$ cannot be countable, a contradiction.