ag.algebraic geometry - Why can't subvarieties separate?

I'm posting my answer to this question as its own question:

Let $V$ be an irreducible projective variety over $mathbb{C}$. Let $U$ be a Zariski open set in $V$. I'll use $V(mathbb{C})$ and $U(mathbb{C})$ to mean $V$ and $U$ equipped with their Euclidean topologies, respectively.

What is the easiest proof that $U(mathbb{C})$ is connected?

Here's the proof I know: Suppose that $U(mathbb{C})$ can be written as a disjoint union of two open sets $A$ and $B$. Since the complement of $U$ in $V$ is a variety of smaller dimension than $V$, a theorem of Remmert and Stein implies that the closures $overline{A}$ and $overline{B}$ of $A$ and $B$ in $V(mathbb{C})$ are projective analytic sets. By Chow's theorem that projective analytic sets are algebraic, $overline{A}$ and $overline{B}$ are subvarieties of $V$. Since they're proper, $V$ is not irreducible, and we have a contradiction.

I guess I'm really asking for the most elementary argument, as I think the above argument is nice intuitively. A reference would be fine.

(To avoid going through the same discussion in the comments that happened at the other question, let me point out that I am aware that irreducible varieties are connected and that $U$ is itself a variety in the sense that it is locally affine. It is just not obvious to me that it is irreducible (without appealing to the above argument).)