Pick a closed convex cone $C\subseteq \mathbf{R}^n$, for some positive integer $n$, such that $C$ is not a vector space. Is it true that $C\cap (-C)=\{0\}$?
Certainly, $D:=C\cap (-C)$ is a closed vector subspace of $\mathbf{R}^n$, and without loss of generality we can assume that $\mathrm{dim}(\mathrm{span}(C))=n$. Now, if the answer were positive, suppose that there exists a nonzero $x_0 \in D$ (hence $-x_0 \in D$). Since $C$ is not a vector space, there exists $x_1 \in C$ with $-x_1\notin C$, i.e., $x_1 \in C\setminus D$. Since $C$ is convex that $\alpha x_0+(1-\alpha)x_1 \in C$ and $-\alpha x_0+(1-\alpha)x_1 \in C$ for all $\alpha \in [0,1]$. Hence the convex hull of $\{x_0,-x_0,x_1\}$ is contained in $C$. How to continue?
That is wrong, take for instance $C=[0,\infty[\times\mathbb R$, then $C\cap -C = \{ 0\} \times \mathbb R$. In general $C\cap -C$ is the largest vector space contained in $C$ (and is empty if $0\notin C$).