I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Show that $u$ is quasi-concave iff, for every $r ∈ \mathbb{R}$, the upper contour set ${x \in X : u(x) ≥ r}$ is convex.
My Proof:
Suppose $u$ is quasi-concave. Then define its upper contour set $U$ for some $r \in \mathbb{R}$. For some $x,y \in U$ s.t. $u(x) \ge u(y)$. Since $u$ is quasi-concave, we know:
$u(\alpha x+(1-\alpha)y) \ge u(y) \ge r$
for some $\alpha \in (0,1)$. So $u(\alpha x + (1 - \alpha)y) \ge r$ and $\alpha x + (1-\alpha)y \in U$, so $U$ is convex.
Now suppose $U$ is convex for any corresponding $r$. Choose $x,y \in X$ s.t. $u(x) \ge u(y)$. $y \in \partial U_r$ for the $r$ s.t. $u(y) = r$. We can also then see that $x \in U_r$. Since $U_r$ is convex, $\alpha x + (1-\alpha)y \in U_r$ for some $\alpha = (0,1)$, and then:
$u(\alpha x + (1-\alpha)y) \ge r = u(y)$
So $u$ is quasi-concave.