A problem in Sigma algebra
Since $\bf{X}$ is a $\sigma-$algebra, $B_1 \equiv X \in \bf X$. Since $\bf{X}$ is infinite, note any subset $B\in \bf X$ st $\emptyset \subset B \subset X$ has the property, since $\bf X$ is infinite, that either $\{B \cap A | A \in \bf X\}$ or $\{B^c \cap A | A \in \bf{X}\}$ must be an infinite subcollection of $\bf X$.
Now choose $B_n, n \geq 2$ in such a way that $B_n$ intersects an infinite subcollection of $\bf X$, and $B_n \subset B_{n-1}$, and $B_n=B_{n-1} \cap A$ for nonempty $A \in \bf X$, $A \not = B_{n-1}$. In this way, we can get a chain $$\ldots \subset B_{n+1} \subset B_n \subset B_{n-1} \subset \ldots$$
of nonempty subsets of $ X$, and hence construct a sequence $$A_1 = B_1 - B_2, A_2 = B_2 - B_3, A_3 = \ldots$$ of mutually disjoint nonempty elements of $\bf X$.
Let $\phi: \bf X $ $ -> [0,1]$ be such that $\phi \big(\bigcup_{n\in N} A_n \big) = \sum_{n \in N} \frac{1}{2^n}$ for any subset $N \subseteq \mathbb{N}$.
Since the sets $\bigcup_{n \in N}A_n$ are distinct for distinct $N \subseteq \mathbb{N}$ by construction, $\phi$ is well-defined. Given $x \in [0,1], $ we have that $x= \sum_{m\in M} \frac{1}{2^m}$ for some $M \in \mathbb{N}$ hence $\phi(\bigcup_{m\in M} A_m)=x$, hence $\phi$ maps $\bf X$ onto an uncountable set. Hence $\bf X$ must be uncountable.
A ($\sigma$-)algebra is a boolean algebra (see here). In particular, it is an $\Bbb F_2$ vector space. Now if it is infinite, then it must be of infinite dimension. It is follows that it contains an uncountable set.
Edit to reflect comments from @ErikWong and @Neel
Note that an infinite dimension vector space need not be uncountable ($\bigoplus_{n=1}^\infty \mathbb F_2$). However, the argument works for $\sigma$-algebras. More precisely, if $S=\{x_1,x_2,\dots\}$ is a (countable) linearly independent subset of a $\sigma$-algebra $A$, then the span of $S$ (under the countable sum) is a subset of $A$ and has cardinal $2^{\mathbb N}$.
Also note that $S$ can be chosen as $\{A_1, A_2, \dots\}$ in Darin's answer.