Lemma: If A and B are disjoint convex sets in a linear space X, then there are complementary convex sets C containing A and D containing B.
I am genuinely confused by this proof given in the book Geometric Functional Analysis by R.B. Holmes. I'm struggling to understand how [u, v] would intersect co({p,q,z}) here. If the notation isn't clear, co denotes convex hull.
The case when $X$ is two-dimensional is almost fully demonstrated by a picture:
(The point $R$ and line connecting to $Y$ was for an argument I was planning on making, but have abandoned. But I don't want to upload a different picture just to remove them.)
However, you also have to examine what happens when $p$ and $q$ are on opposite sides of the line through $x$ and $z$. I'll leave that to you.
When $X$ has more than two dimensions, all $7$ of the points must still lie in the same $3$-dimensional space. But even when $\overline{px}$ and $\overline {qz}$ are skew with each other, $\triangle pyq$ interposes itself between $u$ and $v$.
Think of the whole shape as being made of sheetmetal. Consider $(p,q)$ to be vertical. If $x$ is not in the plane of $p, q, y$, then there is a bend along $(p,y)$. But since $(x,z)$ is straight, there has to be an identical bend along $(q,y)$ in the opposite direction. $(p,x)$ and $(q,z)$ lie on opposite sides of the plane through $p, q, y$ and also on opposite sides of the horizontal plane through $y$. Because of this, any segment connecting a point of $(p,x)$ to a point of $(q,z)$ must pass through the solid triangle $\triangle pyq$. $u$ and $v$ are such points.