Problem: Given $C$ is a closed convex set in E (Euclidean space in $\mathbb{R}^n$). Consider the distance function $$d_C(x) = \inf\{\Vert x-c \Vert : c\in C\}$$ Verify that the proximal operator of $d_C^2$ is $$\dfrac{1}{2}(P_C(x)+x),$$ where $P_C(x)$ is the projection of $x$ on to $C$.
My attempt: Firstly, we have $$\text{prox}_{d_C^2}(x) = (\text{Id} + \partial d_C^2)^{-1}(x).$$ Consider function \begin{align*} (\text{Id} + \partial d_C^2)(u)& = u + \partial d_C^2(u). \tag{1} \end{align*} Since $d_C^2$ is differentiable then \begin{align*} \partial d_C^2(u)& = \nabla d_C^2(u) = 2u - 2P_C(u) \tag{2} \end{align*} From (1) and (2), we claim that $$(\text{Id}+\partial d_C^2)(u) = 3u - 2P_C(u).$$ So, for $x \in E$ we have $$(\text{Id} + \partial d_C^2)^{-1}(x) = \dfrac{1}{3}\left(x+2P_C(u)\right)$$ Therefore $$\text{prox}_{d_C^2}(x) = \dfrac{1}{3}(x + 2P_C(u)).$$ I stuck here since I can make $\text{prox}d_C^2$ to a function of only variable $x$. May you tell me why I'm wrong and how to fix it.
Actually, your expression seems to be the proximal operator to $\frac12 d_C^2$.
For $x \in E$ let $u := (x + P_C(x))/2$. In order to verify $u = \operatorname{prox}_{\frac12 d_C^2}(x)$, we have to check $$ x \in u + \partial \left(\frac12 d_C^2\right)(u).$$ By using $P_C(u) = P_C(x)$, this is easy to verify.