Showing the join of two disjoint projective varieties is a projective variety. I'm looking at the following proposition in Harris' Algebraic Geometry: a first course:

Proposition: Let $X,Y\subset\mathbb P^n$ be disjoint projective varieties. Then the join $J(X,Y)$ of $X$ and $Y$, i.e. the union of all lines in $\Bbb P^n$ which intersect $X$ and $Y$, is a projective variety.

What I'm having trouble with is determining where the disjoint assumption is used. The proof given is as follows (modulo some rewording):

Proof: By [previous example], the locus $\mathscr C_1(X)$ of lines in $\Bbb P^n$ that intersect $X$ is a closed subvariety of the Grassmannian $\Bbb G(1,n)$, and similarly $\mathscr C_1(Y)$ is a closed subvariety of $\Bbb G(1,n)$, so their intersection $Z:=\mathscr C_1(X)\cap\mathscr C_1(Y)$ is closed in $\Bbb G(1,n)$. But then by [previous proposition], the union in $\Bbb P^n$ of all elements of $Z$ is a closed subvariety of $\Bbb P^n$, and this is exactly equal to $J(X,Y)$.

Thanks in advance to anybody who can point out whatever it is I'm missing.

From my reading, it appears that the assumption of disjointness is used in defining the "join" construction, not the proof of showing that it is projective. Indeed, if $X\cap Y$ is nonempty, then the union of all lines in $\Bbb P^n$ which intersect $X$ and $Y$ is exactly $\Bbb P^n$, which is a boring construction.
