Say a $\Bbbk$-algebra is separable if $L\otimes _\Bbbk A$ is reduced for every extension $L/\Bbbk$. Say it's étale if there's an extension $L/\Bbbk$ such that $L\otimes_\Bbbk A\cong \prod_1^nL$.
Here's a two line proof that f.d. separable implies étale:
Let $\overline \Bbbk$ be an algebraic closure of $\Bbbk$. Then $\overline \Bbbk \otimes _\Bbbk A$ is a reduced f.d. $\overline\Bbbk$-algebra. Being finite dimensional reduced implies being Artin so Artin-Wedderburn gives a desired isomorphism.
Is it really necessary to use $\overline \Bbbk$? Why not just take some $L/\Bbbk$ such that $L\otimes _\Bbbk A$ is reduced? In other words, what is the problem with the following:
Let $L/\Bbbk$ be an extension such that $L \otimes _\Bbbk A$ is reduced. This is a reduced f.d. (?) algebra, so Artin-Wedderburn and done.
Update: Thanks to user26857's comments I understand I've been careless. It seems then that for algebraically closed fields the only possible decomposition can be by copies of the field itself. Why is this?