The proximal map of an indicator is Euclidean projection

Since $C$ is nonempty, some $x\in C$ has ${\iota }_C(x)=0<+\mathrm{\infty }$, so the indicator is proper. If $x,y\in C$, convexity of $C$ puts every convex combination in $C$, where the indicator is zero; if either endpoint is outside $C$, the convexity inequality is automatic for an interior coefficient, and at endpoints it is equality. Thus ${\iota }_C$ is convex.

Its epigraph is exactly $C\times [0,+\mathrm{\infty })$: for $x\in C$ the inequality $0\le s$ is required, and for $x\mathrm{\notin }C$ no real $s$ satisfies $+\mathrm{\infty }\le s$. This product is closed because both factors are closed, so ${\iota }_C$ is closed.

If $n=0$, then ${\mathbb{R}}^0=\{0\}$ by [F6], and nonempty $C\subseteq {\mathbb{R}}^0$ is $\{0\}$. The proximal objective has this sole point as its unique minimizer, and it is also the unique nearest point in $C$ to $v=0$. Hence the claim holds for $n=0$.

Now assume $n\ge 1$. The proximal objective is finite exactly on $C$, where it equals $\mathrm{\parallel }u-v{\mathrm{\parallel }}_2^2\mathrm{/}(2t)$. Since $2t>0$, minimizing it over all ${\mathbb{R}}^n$ is equivalent to minimizing distance from $v$ over $C$. By [F4] the latter has a unique minimizer, which is therefore exactly the singleton in [F5].