Given a closed convex set $C \subset E$ (Euclidean space contained in $\mathbb{R}^n$). Prove that $$\text{prox}_{d_C}(x) = x+\min\left\{\dfrac{1}{d_C(x)}, 1\right\}(P_C(x)-x),$$ where $d_C$ is the distance from a point in $\mathbb{R}^n$ to $C$ and $P_C(x)$ is the projection of $x$ onto $C$ defined by $$P_C(x) = \text{argmin}\left\{\Vert x-c \Vert: c \in C\right\}.$$
My attempt: Consider $$\text{prox}_{d_C}(x) = \text{argmin}\left\{d_C(u) + \dfrac{1}{2}\Vert u-x \Vert^2: u \in E\right\}.$$
Case 1: If $ u \in C \subset E$ then $$\text{prox}_{d_C}(x)=\text{argmin}\left\{\dfrac{1}{2}\Vert u -x\Vert^2: u \in C\right\} = P_C(x).$$
Case 2: If $u \in E \setminus C$. Consider the function $g(u) = d_C(u) + \dfrac{1}{2}\Vert u -x\Vert^2$. Since $d_C$ and $\Vert u-x \Vert^2$ are convex then $g$ is convex. Therefore a point $u$ is global minimizer of $g$ if and only if $$\nabla g(u) = (u_1-x_1, \ldots, u_n-x_n)^T = 0.$$ This is equivalent to $u=x$.
I wonder there is something that I do wrong yields to the different result.