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.

You need to do just a little more work: it’s possible that $C$ and $L$ intersect in a single point; of course in that case $C\cap L$ is trivially convex, so let’s assume now that there are at least two distinct points in $C\cap L$, say $p$ and $q$. Then you can argue pretty much as you did, though it could be phrased more clearly.
Added: Here’s one slightly smoother way to phrase your argument.
Suppose that $p$ and $q$ are distinct points of $C\cap L$ and that $\alpha\in\mathbb{R}$ satisfies $0\le \alpha \le 1$; we must show that $\alpha p + (1-\alpha)q \in C\cap L$. Certainly $\alpha p + (1-\alpha)q \in C$, since $p,q\in C$ and $C$ is convex. Moreover, we know that $L = \{tp+(1-t)q:t\in\mathbb{R}\}$, so $\alpha p + (1-\alpha)q \in L$: just take $t=\alpha$. Thus, $\alpha p + (1-\alpha)q \in C\cap L$, as desired, and $C\cap L$ is convex.
