If $S$ is an infinite $\sigma$ algebra on $X$ then $S$ is not countable

1. Because $B_x$ is supposed to be the intersection of all measurable sets containing $x$, but you've found a measurable set containing $x$ strictly inside $B_x$.

2. Because for any measurable set $T$, we have $T=\bigcup_{x\in T}B_x$. Thus, if there are $n$ distinct sets of the form $B_x$, then there are at most $2^n$ elements of $S$.

One can actually change the argument of @Josef slightly to obtain a correct proof. This is probably also what @ncmathsadist is hinting at, although he does not indicate how to obtain the countably infinite collection of disjoint subsets.

Incidentially, the argument below even shows that an infinite $\sigma$-algebra is not only uncountable, but it has at least the cardinality of the continuum.

Let $(A_n)_{n \in \Bbb{N}}$ be a sequence of pairwise distinct (not necessarily disjoint) sets in $S$.

For an arbitrary subset $M \subset X$ let $M^1 := M$ and $M^{-1} := M^c$.

For a sequence $\omega = (\omega_n)_{n} \in \{\pm 1\}^\Bbb{N}$, define

$$ B^\omega := \bigcap_{n \in \Bbb{N}} A_n^{\omega_n}. $$

Observe that $B^\omega \in A$, because it is a countable intersection of elements of $A$.

ALso note that $B^\omega \cap B^\gamma = \emptyset$ for $\gamma = (\gamma_n)_n \neq \omega$, because there is some $n$ such that $\gamma_n \neq \omega_n$, so that

$$ B^\omega \cap B^\gamma \subset A_n^{\omega_n} \cap A_n^{\gamma_n} = \emptyset. $$

Finally, note that for $n \in \Bbb{N}$ arbitrary, we have (why?)

$$ A_n = \bigcup_{\omega \text{ with } \omega_n = 1} B^\omega . $$

Hence, if the set $\{B^\omega \mid \omega \in \{\pm 1\}^\Bbb{N} \}$ were finite, it would easily follow that there could only be finitely many distinct $A_n$, a contradiction.

Hence, there is an infinite family $\omega^{(n)}$ with $B^{\omega^{(n)}} \neq B^{\omega^{(m)}}$ for $n \neq m$.

Let $C_n := B^{\omega^{(n)}}$. By discarding (at most) one set, we can assume $C_n \neq \emptyset$ for all $n$. Using the pairwise disjointness of the $C_n \neq \emptyset$, it is now easy to see that the map

$$ \Gamma : \{0,1\}^\Bbb{N} \to A, (\alpha_n) \mapsto \biguplus_{n \text{ with } \alpha_n = 1} C_n $$

is injective, because

$$ \Theta : A \to \{0,1\}^\Bbb{N}, M \mapsto \left(n \mapsto \begin{cases} 1, & \text{if }M\cap C_{n}\neq\emptyset\\ 0, & \text{if }M\cap C_{n}=\emptyset \end{cases} \right) $$

is a left inverse for $\Gamma$.

Proof by picture:

Take a collection of finitely many disjoint elements of the sigma algebra. Then the picture below shows gives some intuition as to why one can always add a element to this set of pairwise dijoint sets.

Once you have a infinite collection of pairwise disjoint sets one can identify each of these as distinct elements where unions of sets are also distinct. So by taking all countable unions on this collection one would generate a set with uncountably many elements as it is the power set of a countable set.

Show that if a $\sigma$-algebra is infinite, that it contains a countably infinite collection of disjoint subsets. An immediate consequence is that the $\sigma$-algebra is uncountable.

I elaborate ncmathsadist's answer a little bit:

We define the notion of atom: An atom is a non-empty minimal measurable set. That is, a non-empty measurable set $A$ is an atom iff for any measurable set $B\subseteq A$, either $B=\emptyset$ or $B=A$. Note that any two distinct atoms are disjoint. We have the following cases:

Case 1: There are finitely many atoms $A_{1},A_{2},\ldots,A_{n}$ and $\bigcup_{i}A_{i}=X$. In this case, the $\sigma$-algebra is finite and every measurable set is a union of atoms.

Case 2: There are finitely many atoms $A_{1},A_{2}\ldots,A_{n}$ but $\bigcup_{i}A_{i}\neq X$. (This includes the case that there is no atom) In this case, let $Y=X\setminus\bigcup_{i}A_{i}$. We can construct inductively a sequence of measurable set $\{B_{n}\}$ such that $Y=B_{1}\supsetneq B_{2}\supsetneq B_{2}\supsetneq\ldots$. Define $C_{1}=B_{1}\setminus B_{2}$, $C_{2}=B_{2}\setminus B_{3}$ etc, then $\{C_{n}\}$ is a sequence of pairwisely disjoint non-empty measurable sets.

Case 3: There are infinitely many atoms. Choose a countably infinite sub-family of atoms, then they are pairwisely disjoint.