A complete space without isolated points has at least continuum cardinality. At least if you agree to use (some form of) Axiom of Choice.
Choose two disjoint closed balls $B_1$ and $B_2$. Inside $B_1$, choose disjoint closed balls $B_{11}$ and $B_{12}$. Inside $B_2$, choose disjoint closed balls $B_{21}$ and $B_{22}$. And so on. At $n$th step, you have $2^n$ disjoint balls indexed by binary words of length $n$, and you choose two disjoint balls of level $n+1$ inside each ball of level $n$. This is possible because the balls are not single points. Make sure that radii go to zero. Now you have continuum of sequences of nested balls each having a common point.
I'm really not used to thinking about this sort of question (which is why I'm giving it a shot...) but here goes.
Given a $\sigma$-algebra $A$ of subsets of a set $X$, assume that $A$ is infinite. Then $A$ is a poset, with inclusion as the order relation. Apply Zorn's lemma to the poset $P$ of all linearly ordered subsets of $A$ to conclude that there is a maximal linearly ordered subset of $A$; since $A$ is infinite this implies there is an infinite chain $S_1 \subset S_2 \subset \cdots$ of elements of $A$. The pairwise differences $S_{i+1}-S_{i}$ are pairwise disjoint and generate a $\sigma$-subalgebra of $A$ of size at least the size of the power set of $\mathbb{Z}_{\geq 0}$. So any infinite $\sigma$-algebra is at least that big.
Edit: Actually, this combined Matthew's argument below almost finishes the problem: if the cardinality of X is at least that of the power set of $\mathbb{Z}_{\geq 0}$, then the $\sigma$-algebra of countable or cocountable sets in $X$ has cardinality $X$ if $X$ is not of countable cofinality. So a cardinal number is the cardinality of a sigma algebra exactly if it is at least the cardinality of the continuum (no CH needed) and not of countable cofinality.
Best Answer
No, the statement cannot be proven in ZFC without assuming continuum hypothesis or something similar. In fact, it is equivalent to the statement that there are finitely many cardinalities between $\aleph_0$ and $2^{\aleph_0}$, so it is strictly weaker than the continuum hypothesis.
Suppose that there were infinitely many such cardinalities, then you can let $S=\bigcup_{n=1}^\infty S_n$ where $S_n\subseteq(0,1/n)$ has cardinality $\aleph_n$ to obtain a contradition.
On the other hand, if there are only finitely many such cardinalities, then $f(x)=\vert S\cap(-\infty,x)\vert$ must achieve its maximum, say $\aleph_n$ ($n > 0$). If $x_0$ is the infimum of the $x\in\mathbb{R}$ such that $f(x)=\aleph_n$ then $S\cap(x_0,\infty)$ has cardinality $\aleph_n$. Choosing $y_k\in\mathbb{R}$ decreasing to $x_0$, the cardinality of $S\cap(y_k,\infty)$ must be $\aleph_n$ for large enough $k$, otherwise $S\cap(x_0,\infty)=\bigcup_k(S\cap(y_k,\infty))$ is a countable union of sets of cardinality less than $\aleph_n$, so is of cardinality less than $\aleph_n$, giving a contradiction. So, $S\cap(-\infty,y_k)$ and $S\cap(y_k,\infty)$ are both of cardinality $\aleph_n$ for $k$ large enough.