I'm wondering if the following is sufficient to show that for a closed, convex set $S$ $d\left(\mathbf{x},S\right)=\displaystyle\min_{\mathbf{y}\in{S}}\|\mathbf{x}-\mathbf{y}\|$ is convex.
Definition of convexity: $f\left(\theta\mathbf{x} + \left(1-\theta\right)\mathbf{y}\right)\leq\theta f\left(\mathbf{x}\right) + \left(1-\theta\right)f\left(\mathbf{y}\right),\,\,\forall \mathbf{x}\in \mathbb{R}^{n},\,\,\mathbf{y}\in{\mathrm{dom}\left(f\right)}$
Convexity for this function: $\min_{\mathbf{z}\in{S}}\,\|\left(\theta\mathbf{x} + \left(1-\theta\right)\mathbf{y}\right)-\mathbf{z}\|\leq\theta\cdot\min_{\mathbf{y}\in{S}}\,\|\mathbf{x}-\mathbf{y}\| + \left(1-\theta\right)\cdot\min_{\mathbf{z}\in{S}}\,\|\mathbf{y}-\mathbf{z}\|,\,\,\,\theta\in\left[0,1\right]$
Let $\theta=0$, then the above equation reduces to $\min_{\mathbf{z}\in{S}}\,\|\mathbf{y}-\mathbf{z}\|\leq\min_{\mathbf{z}\in{S}}\,\|\mathbf{y}-\mathbf{z}\|, $ which is true.
Let $\theta=1$, then it reduces to $\min_{\mathbf{z}\in{S}}\,\|\mathbf{x}-\mathbf{z}\|\leq\min_{\mathbf{y}\in{S}}\,\|\mathbf{x}-\mathbf{y}\|.$
Are these two facts alone enough to state that $d$ is convex?