I feel blocked with this claim - it sounds intuitively true, just thinking as a jellyfish entering a real line, the intersection of her legs with the real line is certainly finite since the jellyfish is compact - but I stuck with why.
$X$ and $Z$ are closed submanifolds inside $Y$ with complementary dimension. If at lease one of them, say $X$, is compact, and $X \pitchfork Z$, then $X \cap Z$ must be a finite set of points.
I understand that $X \cap Z$ is a zero-dimensional manifold. So it must be a series of disjoint points.
Then I start to guess: this conclusion is perhaps related to each sequence in a compact set has finite subsequence? So let $X \cap Z$ be the sequence, and hence it needs to be finite?
The statement is from Guillemin and Pollack's Differential Topology.
AAs you say, $X\cap Z$ is zero dimensional. If $X$ is compact and $Z$ is closed, then $X\cap Z$ is a closed set in $X$. If it is not finite, then there is a point $x\in X\cap Z$ and a sequence $(x_n)_{n\geq1}$ with values in $X\cap Z$ and all distinct from $x$ such that $x_n\to x$ as $n\to\infty$.
Can you reach a contradition from this?
Notice that $x$ is a point of transverse intersection, so you know how the whole thing is near $x$.