So as a homework I need to prove the following:
Theorem: Show that the set $\mathcal{C}\subseteq\mathbb{R}^n$ is convex if and only if its intersection with the line $L:=\{x=\mathbb{R}^n\,|\, \mathbf{u}+t\mathbf{v},\, t\in\mathbb{R}\}$ is convex, for any $\mathbf{u},\mathbf{v}\in\mathbb{R}^n$.
My sketch proof:
$"\Longrightarrow"$ direction: Assume $\mathcal{C}$ is convex; the line $L$ is convex by definition. Thus $\mathcal{C}\cap L$ is convex since it is the intersection between two convex sets.
$"\Longleftarrow"$ direction: I prove the contrapositive of this claim, i.e. if the set $\mathcal{C}$ is nonconvex, then there exists at least a line $L_*$ such that $\mathcal{C}\cap L_*$ is nonconvex. I then construct an example by drawing a picture to verify the existence of such lines, and finally conclude the proof. $\bigtriangleup$
Here is my issue: I learnt that we can prove existence by constructing an example. However, I am not sure if my mere drawing of a line $L_*$ cutting through a nonconvex set $\mathcal{C}$ (without giving any extra equation describing either set) qualifies as a proof.
It doesn't qualify as a proof. However, think about the definition of ${\cal C}$ being convex: It means that for any $\theta \in [0,1]$ and any $x,y \in {\cal C}$, you should have $\theta x + (1-\theta) y \in {\cal C}$.
$x$ and $y$ are in the intersection of ${\cal C}$ with the line passing through $x$ and $y$. Since this intersection is convex, you know that $\theta x + (1-\theta) y$ is in the intersection. And if it is in the intersection, it is in ${\cal C}$.