[proofplan] Let $\alpha$ denote the supremum of $u$ on $C$. The exposed set is exactly the intersection of $C$ with the affine hyperplane where $u=\alpha$, so convexity follows from linearity of $u$ and convexity of $C$. For the facial property, if a strict convex combination lies in this level set, then the weighted average of the two numbers $u(x)$ and $u(y)$ equals their common upper bound $\alpha$; since each is at most $\alpha$, both must equal $\alpha$. [/proofplan] [step:Fix the supporting level attained by the exposed set] Since $F(C,u) \neq \varnothing$, choose $x_0 \in F(C,u)$. Define \begin{align*} \alpha := \sup_{z \in C} u(z). \end{align*} Because $x_0 \in F(C,u)$, we have $u(x_0)=\alpha$. In particular, $\alpha \in \mathbb{R}$, and for every $z \in C$, \begin{align*} u(z) \leq \alpha. \end{align*} Thus \begin{align*} F(C,u)=\{x \in C : u(x)=\alpha\}. \end{align*} [/step] [step:Prove that the exposed set is convex] Let $a,b \in F(C,u)$ and let $s \in [0,1]$. Since $C$ is convex, the point \begin{align*} c := sa+(1-s)b \end{align*} belongs to $C$. Since $u:\mathbb{R}^n \to \mathbb{R}$ is linear, \begin{align*} u(c) &= u(sa+(1-s)b) \\ &= s u(a)+(1-s)u(b) \\ &= s\alpha+(1-s)\alpha \\ &= \alpha. \end{align*} Therefore $c \in F(C,u)$. Hence $F(C,u)$ is convex. [/step] [step:Show that any segment whose interior point is exposed has exposed endpoints] Let $x,y \in C$ and $t \in (0,1)$ satisfy \begin{align*} w := tx+(1-t)y \in F(C,u). \end{align*} Since $w \in F(C,u)$, we have $u(w)=\alpha$. By linearity of $u$, \begin{align*} \alpha = u(w) = u(tx+(1-t)y) = t u(x)+(1-t)u(y). \end{align*} Because $x,y \in C$, the definition of $\alpha$ gives \begin{align*} u(x) \leq \alpha, \qquad u(y) \leq \alpha. \end{align*} Suppose first that $u(x)<\alpha$. Since $t>0$ and $1-t>0$, and since $u(y)\leq \alpha$, \begin{align*} t u(x)+(1-t)u(y) &< t\alpha+(1-t)\alpha \\ &= \alpha, \end{align*} contradicting $t u(x)+(1-t)u(y)=\alpha$. Hence $u(x)=\alpha$. The same argument with $x$ and $y$ interchanged gives $u(y)=\alpha$. Therefore $x,y \in F(C,u)$. [/step] [step:Conclude that the exposed set is a face] We have proved that $F(C,u)$ is convex and that whenever $x,y \in C$ and $t \in (0,1)$ satisfy $tx+(1-t)y \in F(C,u)$, both endpoints $x$ and $y$ belong to $F(C,u)$. These are exactly the defining properties of a face of $C$. Therefore $F(C,u)$ is a face of $C$. [/step]