There are already two answers pointing out why your statement cannot hold as stated, so let's see if we can fix it.
Let $X, Y\subseteq \mathbb P^N$ be two irreducible (quasi-projective) algebraic varieties of dimension $k$ and $l$ respectively. Then $\overline X,\overline Y\subseteq \mathbb P^N$ are two closed irreducible algebraic varieties of dimension $k$ and $l$ respectively. By the Projective Dimension Theorem you obtain that
Every irreducible component of the intersection $\overline X\cap\overline Y$ has dimension at least $k+l-N$.
This implies that if your initial $X$ and $Y$ are disjoint, then your desired statement cannot hold.
On the other hand since you assumed that $X$ and $Y$ intersect transversally, basically you only need to worry about the complements, that is, the interesting intersections are $\overline X\cap (\overline Y\setminus Y)$ and $(\overline X\setminus X)\cap \overline Y$.
If you know that these intersections are transversal and $X\cap Y$ is not empty, then I think what you want follows.
A perhaps interesting consequence of this is that if those intersections are transversal, then $X\cap Y\neq \emptyset$.