Let $V,U$ vector subspaces of $\mathbb{C}^{n+1}$ of dimension $r+1$, $s+1$.
Let $\pi : \mathbb{C}^{n+1}/\{0\} \longrightarrow \mathbb{CP}^n$ the projection map to complex projective space defined by $\pi:(x_0, ... ,x_n)\mapsto(x_0:...:x_n)$.
This is a question that came up while doing homework and I wanted to check some reasoning. It is useful to have that $\pi(V)\cap\pi(U) = \pi(V\cap U)$ for a homework question. Is this true? It seemed to be true based on my currently poor intuition about projective space, so I tried to prove it.
My reasoning for a simple proof was to prove both inclusions. Specifically, if $u\in \pi(U\cap V)$ then there is some $x\in U\cap V$ st $u=\pi(x)$. But then $x$ is in both $U$ and $V$ so $u\in \pi(U)\cap\pi(V)$.
Going in the other direction, if $u\in \pi(U)\cap\pi(V)$ then there is $x\in U$ and $y\in V$ such that $\pi(x)=\pi(y) = u$. But if this is true then $x$ is a scalar multiple of $y$ and vis versa (by the properties of vector subspaces), so $x\in U\cap V$. Thus $u\in\pi(U\cap V)$ and we have proved the sets are equivalent.
Yes, this is true, and your reasons are correct. Both describe the set of one dimensional subspaces minus the origin contained in both $U$ and $V$. Since fibers of $\pi$ are one dimensional subspaces minus the origin, any fiber that intersects a subspace must be contained in that subspace. Note that $f(A\cap B)\subseteq f(A)\cap f(B)$ is always true for the same reason you gave that $\pi(U\cap V)\subseteq\pi(U)\cap\pi(V)$.