Convex sets proof

Well, first note that if we only have two points $x_1$ and $x_2$, then all that's being said is whenever $a + b = 1$ the point $a*x_1 + b*x_2$ is in $D$. This is very clear though, because $b = 1-a$ and so the point in question is $a*x_1 + (1-a)*x_2$, which is a point on the line between $x_1$ and $x_2$.

Generally speaking, if we have points $x_1, ..., x_k$, and $\sum_{i=1}^k a_i = 1$, then you can write $a_1 + ... + a_{k-1} = 1 - a_k$ to get that

$\sum_{i=1}^k a_i x_i = a_k x_k + (1-a_k)\sum_{i=1}^{k-1} \frac{a_i}{1 - a_k} x_k $

The points $x_k$ and $\sum_{i=1}^{k-1} \frac{a_i}{1 - a_k} x_k$ may by induction be assumed to be points in $D$, so this forms the induction step of the proof.

You can proceed by induction on $k$, the case $k=1$ being trivial. If $k>1$, let $u=\sum_{i=1}^{k-1}a_i=1-a_{k}$. If $u=0$, then $\sum_{i=1}^k a_ix_i = x_k\in D$. Otherwise let $b_i=\frac{a_i}u$ and observe that $y:=\sum_{i=1}^{k-1}b_ix_i\in D$ by induction assumption because $\sum_{i=1}^{k-1}b_i=1$ and all $b_i\ge 0$. Then $$\sum_{i=1}^k a_ix_i = x_k+u(y-x_k)$$ is a point on the line segment from $x_k$ to $y$, hence in $D$.