Show that if $S_1=\{(x_1,x_2):x_1+x_2 \le 0,x_1 \ge 0\}$ and $S_2=\{(x_1,x_2):x_1-x_2 \ge 0,x_1 \le 1\}$ and $S:=S_1 \cup S_2$, then $S_1$ and $S_2$ are convex but $S$ is not.
Take $x=(x_1,x_2),y=(y_1,y_2) \in S_1$ and an arbitrary $\lambda \in [0,1]$, by the definition of $S_1$ we have that $x_1+x_2 \le 0,x_1 \ge 0$ and $y_1+y_2 \le 0,y_1 \ge 0$, non-negativity of $\lambda$ implies that $$\lambda x_1+ \lambda x_2 \le 0,\lambda x_1 \ge 0$$ and $$(1-\lambda)y_1+(1-\lambda)y_2 \le 0,(1-\lambda)y_1 \ge 0$$
Which shows that $\lambda x_1+(1-\lambda)y_1+\lambda x_2+(1-\lambda)y_2 \le 0$ ,$\lambda x_1 \ge 0+(1-\lambda)y_1 \ge 0$
By the definition of $S_1$, $\lambda x+ (1-\lambda)y \in S_1 \implies $ $S_1 $ is convex.
A similar process can be used to show the convexity of $S_2$.
However showing non-convexity of $S=S_1 \cup S_2$ was slighly difficult and so far I could not find $x,y \in S$ and $\lambda \in [0,1]$ such that $\lambda x+ (1-\lambda)y \notin S$
Best Answer
Drawing a picture really helps.
$ (3,-3)\in S_1$, $(1,0) \in S_2$ but $\frac 1 2 (1,0)+\frac 1 2(3,-3) \notin S_1\cup S_2$.