Suppose the polar factorization of $X = YZ$.
Since $X$ is non-singular we can re-write $X = (UV^T)(V\Sigma V^T)$.
Now, question is can you prove that $YZ = ZY$ if $X$ is normal? In other words, can you prove:
$$ (UV^T)(V\Sigma V^T)=(V\Sigma V^T)(UV^T) $$
First, verify that $X^TX = Z^2$ and $XX^T = Y^TZ^2Y$. If $X$ is normal, then we have $$ Z^2 = Y^TZ^2Y \implies YZ^2 = Z^2Y $$ Now, note that $Z$ is the unique positive semidefinite square root of $Z^2$. By taking a close look at the theorem guaranteeing positive semidefinite square roots, note that $$ YZ^2 = Z^2Y \implies YZ = ZY $$ which is the desired conclusion. More specifically, it suffices to note that there exists a polynomial $p$ such that $Z = p(Z^2)$.