$$f(s)=-\left(\min_{w\in C}(w^\top s+\frac12\|w\|_2^2)\right)$$ where $C$ is a compact and convex set in $\mathbb{R}^n$
I believe that the function is concave, but I need to show that it is convex (which could be false). Please give me any result. I am thinking of using this link which answers that the minimum of a parametric convex function is convex to prove that it is concave.
Given any family of convex functions, the pointwise supremum is convex. See the answer here: Prove the supremum of the set of affine functions is convex
(It assumes that the domain is compact, but the proof does not use this, and in any case one can always assume the domain is compact by restricting to a line segment.)
Multiplying by $-1$, we obtain that the pointwise infimum of concave functions is concave.
Affine functions are concave, hence so is the infimum in the definition of $f(s)$, so that $f(s)$ is convex. This argument does not require that $C$ is convex.
Note: it is not automatic that the minimum in your question exists (this can fail for certain $C$ if the term $\frac12 \lVert w \rVert^2$ is not there). But you can write it as $$f(s) = \frac12 \lVert s \rVert^2 - \inf_{w \in C} \frac12 \lVert s + w \rVert^2 $$
and this infimum is attained because $C$ is closed. Moreover, we see that $$f(s) = \frac12 \lVert s \rVert^2 - \frac12 d(s, -C)^2 \,.$$