Let me elaborate on Simon Henry's nice answer.

What we prove is that the Cohen algebra, the completion of the unique countable atomless Boolean algebra, is not isomorphic to any $\sigma$-algebra.

To see this, it suffices to show that every $\sigma$-algebra $\Sigma$ on a set $X$, where there is a countable dense set, has atoms. Fix any $x$, and fix the countable dense set $b_0,b_1,\ldots$ in $\Sigma$, so that every nonzero set contains one of them. Fix any $x\in X$, and let $a_n$ be either $b_n$ or the complement, so that $x\in a_n$. It follows that $\bigwedge_n a_n$ has $x$, but it must be $0$, since the $b_n$'s are dense. Contradiction.

Let me elaborate on Simon Henry's nice answer.

What we prove is that the Cohen algebra, the completion of the unique countable atomless Boolean algebra, is not isomorphic to any $\sigma$-algebra. Since the Cohen algebra has a countable dense set, this follows immediately from the following:

Lemma. If a $\sigma$-algebra $\Sigma$ has a countable dense set, then it must have atoms.

Proof. Enumerate the countable dense set $b_0,b_1,\dots$, where dense here means that every nonempty set in $\Sigma$ must contain some $b_k$ as a subset. Now fix any point $x$ in the underlying set, and let $a_n$ be either $b_n$ or the complement, chosen to ensure $x\in a_n$. Thus, $x\in a^*$, where $a^*=\bigcap_n a_n$ is the intersection. Notice that $a^*$ decides every $b_n$, in the sense that it is contained either in $b_n$ or in the complement. Since it is not empty, it must therefore be an atom. QED