It is true indeed without assuming the Hausdorff separation axiom on the topological vector space  $E$, for the reason that we can quotient over the closure of the origin, $\overline{\{0\}}$.


Let's denote $\pi:E\to \tilde E:=E/\overline{\{0\}}$ the quotient projection, and consider $\tilde K:=\pi (K)$, a compact convex subset of $\tilde E$. 



For any $x\in K$ the set
$T^{-1}(T(x))$ is a nbd of $x$ by assumption, hence it   contains the closure of  $x$ in $K$,  that is all $u\in K$ such that $\pi x=\pi u$. By symmetry we conclude that $T(x)=T(u)$ whenever  $\pi x=\pi u$. So the map $T$ factor through $\pi$, and there is a well-defined map $\tilde T:\tilde K\to 2^{\tilde K}$ such that for any $x\in K$, $\tilde T(\pi x)=\pi(T(x))$. 

The multi $\tilde T$  satisfy the hypotheses of  Browder's fixed point theorem for Hausdorff TVS, because for any $\tilde x\in \tilde K$, $\tilde T(\tilde x)$ is a non-empty convex subset of $\tilde K$ , and for any $y\in E$   
$$\tilde T^{-1}(\pi y)=\bigcup_{v\in \overline y}\pi \big(T^{-1}(v)\big) \, , $$
an open subset of $K$ because $\pi:K\to\tilde K$ is an open map.

In consequence there is $u_0\in K$ such that $\pi u_0\in\pi\big(T(u_0)\big)$. This means that there is $x_0\in K$ such that $\pi u_0=\pi x_0$ and  $x_0\in T(u_0)$: but as observed $T(u_0)=T(x_0)$ and $x_0$ is a fixed point of $T$.