Here is a proposition from Griffith and Harris, Principles of Algebraic Geometry.
Let $V$ be an analytic variety on a complex manifold $M$, let $V^*$ be its points of smooth points. If $V^*$ is connected, then $V$ is irreducible.
The proof in the book is very short. It goes like this:
Suppose $V=V_1\cup V_2$ is a decomposition of $V$. Then $V_1\cap V_2$ is contained in the set of singularities of $V$. Hence $V^*$ can not be connected.
I cannot figure out why $V$ must be singular at the intersection of $V_1$ and $V_2$. This is wrong when the $V_i$'s are allowed to be reducible. For example, $V$ consists of three lines on the plane, $V_1$ is the first two lines and $V_2$ is the last two. But in this case the choice of $V_i$'s has some kind of redundancy.
To make it simple, let's assume that it has precisely two irreducible components and try to derive a contradiction. I am already convinced that, in your notation, $V\cap U$ is irreducible when $V\cap U$ is smooth. What I am afraid of is a picture like this: for example, $V_1$ is the line $y=0$ on the plane and $V_2$ looks like the graph of a bump function. I know this would never happen, because analytic varieties are "rigid". But I don't know how to exclude this possibility.
Can anybody give me a hand? Thanks!