It is not a correct proof. Say that $P(x)$ is the statement "$x$ is a ruby" and $Q(x)$ is the statement "$x$ is red".
Your claim is:
If the LHS is true, then $Q(x)$ must be true for all values of $x$.
The LHS, $\forall x[P(x)\to Q(x)]$ is quite true, since for any $x$, if $x$ is a ruby then it is red. But contrary to your claim, $Q(x)$ is not true for all values of $x$, since not every $x$ is red—crows are not red, and neither is snow.
(However, the RHS is still true; it says that if every $x$ were a ruby, then every $x$ would be red, and this is correct, if the LHS is. But your argument is wrong.)