There is some debate about whether or not the mapping $(X, Y) \mapsto E[XY]$ defines an inner product. If it does, then your result follows from the standard Cauchy-Schwarz inequality in general inner product spaces.
But, even if it's correct, this approach is rather a cop-out, and could easily be regarded as circular reasoning.
To avoid the circular reasoning, just adapt any proof of the Cauchy-Schwarz inequality. A few of them are given on this wikipedia page.
Here is one such proof: As your hint says, we know that $E[(kX+Y)^2] \ge 0$, for any real $k$. Expanding this, we get:
$$k^2E[X^2] + 2kE[XY] + E[Y^2] \ge 0$$
Let's regard this as a quadratic in $k$. It is non-negative for all $k$, so it has at most one real root, so its discriminant (the $b^2 - 4ac$ thing) must be $\le 0$. In other words
$$ (2E[XY])^2 - 4E[X^2]E[Y^2] \le 0$$
and the result follows immediately.
Note that I didn't prove the statement about when equality occurs. In fact, I don't think this statement is true, as WimC's comment pointed out.