Why is the realification of a simple complex Lie algebra a semisimple real Lie algebra?

Actually, a stronger statement holds true: Denoting the "realification" of a complex Lie algebra $L$ by $Res_{mathbb Cvert mathbb R}L$, then

$L$ is simple iff $Res_{mathbb Cvert mathbb R}L$ is simple.

Namely, for the non-trivial direction, say we've already shown that $Res_{mathbb Cvert mathbb R}L$ is semisimple like in José Carlos Santos' answer. Let $S$ be a simple component of $Res_{mathbb Cvert mathbb R}L$. Now for any $0 neq lambda in mathbb C$, we have that $lambda S$ is also an ideal of $Res_{mathbb Cvert mathbb R}L$, and $[S, lambda S] = lambda S neq 0$ hence by simplicity $S subseteq lambda S$. But then we actually have equality $S=lambda S$, because they have the same real dimension. But that means the complex span of $S$ in $L$ is $S$ itself, hence it is a non-zero ideal of $L$, hence all of $L$ by $L$ being simple, and hence as a set it is also all of $Res_{mathbb Cvert mathbb R}L$.

This statement and proof, including the step provided by José Carlos Santos' answer, works far more generally for any finite extension of characteristic $0$ fields $Kvert k$ instead of $mathbb C vert mathbb R$, and I took the proof from Bourbaki's Lie Groups and Algebras, chapter I §6 no. 10.

If one digs deeper into the theory, actually the following holds true: If the Dynkin diagram to $L$ is of type $R$ (i.e. $A_n, B_n, C_n, D_n, E_6, E_7, E_8, F_4$ or $G_2$), then $Res_{mathbb Cvert mathbb R}L$ is a quasi-split real Lie algebra whose Satake diagram consists of two copies of the Dynkin diagram of $L$, with complex conjugation flipping those two copies. That real Lie algebra is simple, but not "absolutely simple", since if one complexifies it again, one gets a direct sum of two copies of the original $L$. Again, this holds true in greater generality, compare section 4.1 (especially p. 67) of my thesis.