Mashke's theorem states:
Let $K$ be a field and let $G$ be a finite group. $K[G]$ is a semisimple ring iff $|G|$ is not divided by $\textrm{char}(K);$ in particular, it's true if $\textrm{char}(K)=0.$
I know a "standard" proof for the general situation. Consider a special case when $K$ is a (not nessecary proper) subfield of $\mathbb C.$ I wonder if there is a kind of simplified proof for the inverse implication (i.e. "from right to left"). I think $G$-invariant dot product on $K[G]$ might be the key; one can consider it as $\Big(\sum\limits_{g\in G}\alpha_gg,\sum\limits_{g\in G}\beta_gg\Big)=\sum\limits_{g\in G}\alpha_g\overline{\beta_g}.$ Obviously, $(u,v)=(gu,gv)$ for all $u,v\in K[G]$ and $g\in G.$ Thus, all the operators of a representation are unitary.
Is it possible to complete this idea to a proof and, if so, how?
Yes. More generally, any finite-dimensional unitary representation $V$ of a group is semisimple. The point is that if $W\subseteq V$ is an invariant subspace, then the orthogonal complement $W^\perp\subseteq V$ is also invariant. The proof is simple: if $x\in W^\perp$ and $y\in W$, then $(gx,y)=(x,g^{-1}y)=0$, so $gx$ is orthogonal to every element of $W$ and is thus in $W^\perp$. So $W$ is a direct summand of $V$ as a $K[G]$-module, with direct complement $W^\perp$.