I am being stupid here. $G$ is a connected algebraic group and $s$ is a semisimple element. Let $T$ be a maximal torus. Then if $T$ is contained in $C_G(s)$, then $s$ is in $T$. I got stuck here. I know semisimple elements are in maximal tori. But here, could $s$ be in a different torus $T'$ conjugate to $T$? I am trying to show the two tori are in fact one.
This question is from P112 of Gunter Malle's "linear algebraic group and finite group of Lie type".
Proposition 14.1) Let $G$ be connected, $s ∈ G$ semisimple, $T ≤ G$ a maximal torus. Then $s ∈ T$ if and only if $T ≤ C_G(s)^◦$. In particular, $s ∈ C_G(s)^◦$.
Proof: As $T$ is abelian, $s ∈ T$ if and only if $T ≤ C_G(s)$, which is equivalent to $T ≤ C_G(s)^◦$ as $T$ is connected.
I'm not quite sure what you're getting at, but I think it seems like maybe you want the following:
Proof: If $s\in T(F)$ then since $T$ is abelian we know that $T$ centralizes $s$ and thus $T\subseteq C_G(s)$. Since $T$ is connected we then also have that $T\subseteq C_G(s)^\circ$.
Remark: Note that the above argument also shows, since every semisimple element of $G(F)$ is in $T(F)$ for some maximal torus $T$ of $G$, that $s\in C_G(s)^\circ$ (it's obvious that $s\in C_G(s)$, but less obvious it's in $C_G(s)^\circ$).
Conversely, suppose that $T$ is a maximal torus of $C_G(s)^\circ$. Since $C_G(s)^\circ$ is reductive (e.g. see [Humphreys, Theorem in §2.2]--it's easy to reduce to the semisimple case) we know that $Z(C_G(s))\subseteq T$ (e.g. see [Milne, Proposition 21.7]). But, since $s\in Z(C_G(s))(F)$ this implies that $s\in T(F)$. $\blacksquare$
References:
[Humphreys] Humphreys, J.E., 2011. Conjugacy classes in semisimple algebraic groups (No. 43). American Mathematical Soc..
[Milne] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.