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 $\tilde E$. From the second assumption on $T$ it follows that for any $x\in K$ and $u\in K$ such that $\pi x=\pi u$, any $y$ belongs to $T(x)$ if and only if it belongs to $T(u)$, that is $T(x)=T(u)$. 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 theorem, 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.
By the above Browder's fixed point theorem for Hausdorff TVS, 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 $T$.