Complete reducibility and field extension Let $\pi$ be a representation of a Lie algebra $L$ in a finite-dimensional linear space $V$ over the field $F$. Let $K$ be a field extension of $F$. Let $\pi_K=\pi\otimes K$ be the corresponding representation of $L_K$ in $V_K$. Why if $\pi_K$ is completely reducible then $\pi$ must be completely reducible too?
 A: It is worth pointing out that there are general statements that semisimplicity can be checked over an algebraic closure. The following statement is used:

Let $k$ be a field and $A$ a $k$-linear abelian category such that
  $End(X)$ is finite dimensional for every object $X$ in $A$. Then $A$
  is semisimple if and only if $End(X)$ is a semisimple $k$-algebra for
  all $X$ in $A$.

(a proof can be e.g. found in Proposition 5.14 of Milne's lecture notes on Lie algebras: https://www.jmilne.org/math/CourseNotes/LAG.pdf)
Furthermore Lemma 5.11 in Milne's notes says that

Let $A$ be a $k$-algebra. If $K \otimes_k A$ is semisimple for some
  field $K$ containing $k$, then $A$ is semisimple; conversely, if $A$
  is semisimple, then $K \otimes_k A$ is semisimple for all fields $K$
  separable over $k$.

Edited after YCor's comment: If one applies these statements to the special case of the category $A = Rep(\mathfrak{g})$ over a field $k$, one gets that if $Rep(\mathfrak{g}_K)$ is semisimple for some extension field $K$ of $k$, then $Rep(\mathfrak{g})$ is semisimple. This answers the initial question. If $V$ is semisimple in $Rep(\mathfrak{g})$, then $V \otimes K$ might not be semisimple, but this holds if either $K$ is separabel over $k$ or if $dim(V) < p$ where $char(k) = p$ (see the comment in Milne's notes - this is originally due to Serre).
A: Amazingly, I cannot see an elementary solution. I believe there should be one.
Otherwise, one can expand the comment of YCor with some standard Ring Theory. Let $A$ be the image of $U(L)$ in $End_F(V)$. Then $A_K = A \otimes_F K$ is the image of $U(L_K)$.
Now suppose $\pi$ is not completely reducible. This means that $A$ is not a semisimple algebra. This means that its Jacobson radical $J$ is non-zero. Since $A$ is finite-dimensional, the Jacobson is nilpotent: $J^n=0$ for some $n>1$. Then $J_K$ is a nilpotent ideal of $A_K$ so that $A_K$ is not a semisimple algebra. Hence, $\pi_K$ is not completely reducible. 
PS Let us dig into the last implication. Suppose $\pi_K$ is completely reducible but the Jacobson $J(A_K)\neq 0$. As $V_K$ is a faithful $A_K$-module, $J(A_K)V_K\neq 0$. By complete reducibility there exists a $A_K$-direct complement $V_K = DirectComplmnt \oplus J(A_K)V_K$. By Nakayama Lemma, $V_K = DirectComplmnt$. Contradiction!!
