Wikipedia has the one I've usually seen.
Cauchy-Schwarz Inequality
Nothing especially tricky about it, just a really clever setup.
If I remember correctly, Cauchy proved the inequality that you don't want proven. The generalisation that you do want proven was proven later by Schwarz. So, I think that the former inequality ought to be called the Cauchy inequality. Oh well.
The proof of the (general) Cauchy-Schwarz inequality essentially comes down to orthogonally decomposing $x$ into a component parallel to $y$ and a component perpendicular to $y$, and using the fact that the perpendicular component has a non-negative square norm. That is, we start from the statement:
$$\left\|x - \frac{\langle x, y \rangle}{\langle y, y \rangle} y\right\|^2 \ge 0.$$
The rest is just expanding the inner product. We have,
\begin{align*}
0 &\le \left\langle x - \frac{\langle x, y \rangle}{\langle y, y \rangle} y, x - \frac{\langle x, y \rangle}{\langle y, y \rangle} y\right\rangle \\
&= \langle x, x \rangle - \frac{\overline{\langle x, y \rangle}}{\langle y, y \rangle} \langle x, y \rangle - \frac{\langle x, y \rangle}{\langle y, y \rangle} \langle y, x \rangle + \frac{\langle x, y \rangle}{\langle y, y \rangle} \frac{\overline{\langle x, y \rangle}}{\langle y, y \rangle} \langle y, y\rangle \\
&= \langle x, x \rangle - 2\frac{|\langle x, y \rangle|^2}{\langle y, y \rangle} + \frac{|\langle x, y \rangle|^2}{\langle y, y \rangle} \\
&= \langle x, x \rangle - \frac{|\langle x, y \rangle|^2}{\langle y, y \rangle}.
\end{align*}
Rearranging yields the inequality.
Best Answer
No it doesn't hold in $L^1$. Take $f(x)=g(x)=\frac{1}{\sqrt{x}}$ for $x \in (0,1)$ and $f(x)=g(x)=0$ elsewhere.
$\Vert f \Vert_1=\Vert g \Vert_1=2$ but $\int fg =+\infty$.