So $U_{1}$, $U_{2}$ are vector subspaces of V. I need to show that the linear subspace P($U_{1}$+$U_{2}$)$\subseteq$ P(V) is the set of points obtained by joining each X$\in$ P($U_{1}$) and Y$\in$ P($U_{2}$) by a projective line.
What is a good way to prove this? Thanks.
Let $ \pi : V \backslash 0 \to \mathbb P(V)$, $v \mapsto [v]$ the projection.
If $U,U'$ are two subspaces, the subspace $U+U'$ is precisely generated by the sums $u + u'$ where $u \in U, u' \in U'$.
Take any line (through the origin) $\ell$ contained in $U+U'$.
$\ell$ can be written as $t\xi$ for some vector $\xi \in U + U'$. We want to show that there is a plane $P$ which contains $\ell$, a line of $U$, says $\text{span}(u)$ and a line of $U'$, says $\text{span}(u')$, so that $\pi(\xi)$ will be on the line between $\pi(u)$ and $\pi(u')$ and such lines will generate $\mathbb P(U+U')$.
If $\xi \in U$ or $\xi \in U'$ that means $\pi(\xi) \in \mathbb P(U)$ or $\pi(\xi) \in \mathbb P(U')$. If $\pi(\xi) \in P(U)$, take $\xi = u$ and $u'$ any point in $U'$, so join this point to any point in $\mathbb P(U')$. Repeat the same construction if $\pi(\xi) \in \mathbb P(U')$. This gives us a line $L$ between $\mathbb P(U)$ and $\mathbb P(U')$ with $\pi(\xi) \in L$.
Now, assume $ \xi = u + u'$. Then, $\xi$ is obviously in the plane generated by $u$ and $u'$. So $\pi(\xi) \in \text{proj} \{\pi(u), \pi(u')\}$ and that's what we wanted.