Convexity of intersection I have been asked to prove that, given a convex set $C$, its intersection with a line is also convex.
From convexity definition, I have that $\forall x_1,x_2\in C, \alpha x_1+\beta x_2 \in C$ with $\alpha,\beta\ge0, \alpha+\beta=1$. 
If I have $x_3,x_4 \in C \cap L$, every point in the line can be expressed as $x = x_4 + t (x_3-x_4)=tx_3+(1-t)x_4$. If the convex combination $\alpha x_3+(1-\alpha) x_4$ belongs to $C$, then it is trivial to show that, just by making $t=\alpha$, it also belongs to L, hence to $C \cap L$.
As always, I think I am oversimplifying or forgetting something. Is my proof right, or am I missing the crucial point?
Thanks in advance.

In general, the intersection of two convex sets is again a convex set. To do this, pick $x,y \in C_1 \cap C_2$ and $\alpha \in (0,1)$. Then by convexity of $C_1$ we have that $\alpha x +(1-\alpha)y \in C_1$. The same argument proves that $\alpha x +(1-\alpha)y \in C_2$ so $\alpha x +(1-\alpha)y \in C_1\cap C_2$ and by definition $C_1 \cap C_2$ is convex.
A line is a convex set, so the intersection of a line and a convex set is again convex.
