Anisotropic algebraic groups have no unipotent elements I have found the following fact stated in a number of places: 
If $k$ is any field, a connected reductive group $G$ is anisotropic if and only if its only unipotent element is $e$ and $\mathrm{Hom}_k(G, \mathrm{G}_m)$ is trivial.
For instance, this appears in Section 3.4 of Springer's Corvallis article. However, I have been unable to track down a reference for a proof of this result.
 A: What you have asked for is indeed a fact in characteristic zero. See  Corollary 8.5 of Borel-Tits (Publ. IHES at Numdam, unrestricted access). 
However, it is false in positive characteristic. See also section 4 of the same article (where other fields are considered). Indeed, the group $\mathrm{PGL}_1(D)$ over a field $k$ of positive characteristic $p$, $D$ a central division algebra over $k$ of degree $p$ , can have "bad" unipotent elements coming from purely inseparable extensions of $k$ lying in $D$. What is true is that $G$ is anisotropic iff it does not have a proper parabolic subgroup defined over $k$.
In any case, this article by Borel and Tits is the standard reference for reductive groups over arbitrary fields.
A: This is true when $k$ is a perfect field. Reference for the proof is the paper by Borel and Tits.
When $k=\mathbb{R}$, it is more familiar situation. Anisotropic reductive $\mathbb{R}$-groups are nothing but the compact Lie groups.
Then by torus theorem, any element is in a maximal torus, which means any element is semisimple.
