Def. Let $G$ be an algebraic group and let $P \subset G$ be a closed subgroup. Then $P$ is said to be parabolic if $G/P$ is complete.
Apparently, we can prove
Lemma. $P \subset G$ is parabolic $\Leftrightarrow$ $(P \cap G^0) \subset G^0$ is parabolic.
so that we needn't assume $G$ to be connected in the definition of "parabolic".
However, I seem to be missing one or another insight. I know that $G^0$ is a subgroup of finite index, and I suspect that $G^0/(P\cap G^0)$ is an irreducible component of $G/P$. Could someone give a hint for how to give a rigorous proof?
(One can draw a diagram like the one used in the nine lemma, but strictly speaking this only holds in an Abelian category or in the category of groups. Of course $0 \to P \to G \to G/P \to 0$ isn't an exact sequence in either category.)
One has an exact sequence of group schemes
$$1\to G^\circ\to G\to\pi_0(G)\to 1$$
where $\pi_0(G)$ is the group of connected components of $G$ (e.g. see [Mil, Pg. 15-16]). Note then that one gets an fibration
$$1\to G^\circ/(P\cap G^\circ)\to G/P\to \pi_0(G)/P\to 1$$
where $\pi_0(G)/P$ denotes the quotient of $\pi_0(G)$ modulo $P$. This implies that $G^\circ/(P\cap G^\circ)$ closed embeds into $G/P$ as one of the components of $G/P$ as desired. In particular, since $G$ acts transitively on the components of $G/P$ we see that $G/P$ is proper if and only if $G^\circ/(P\cap G^\circ)$ is proper, which gives the lemma.
[Mil] Milne, J.S., 2017. Algebraic groups: The theory of group schemes of finite type over a field (Vol. 170). Cambridge University Press.