How is the Lie algebra $\mathfrak{sl}_{2}(\mathbb{C})$ generated by the one element $(E_{11} - E_{22})$? In the article Classification of simple complex Lie algebras, 2nd paragraph of chapter 7 (p.15, bottom), the author considers the basis
$$
x = E_{21} := \left[
\begin{array}{ll}
0 & 0\\
1 & 0\\
\end{array}
\right], \quad
y = E_{12} := \left[
\begin{array}{ll}
0 & 1\\
0 & 0\\
\end{array}
\right], \quad
h = \left[
\begin{array}{lr}
1 & 0\\
0 & -1\\
\end{array}
\right]
$$
for the Lie algebra $\mathfrak{sl}_{2}(\mathbb{C})$.  The following relations hold:
$$
[h, x] = 2x, \quad [h, y] = -2y, \quad [x, y] = h.
$$
The author asserts that $h$ alone generates the entire Lie algebra.  I have tried to carry out the generation and got
$$
hh = I, \quad {1 \over 2}(hh + h) = E_{11} := \left[
\begin{array}{ll}
1 & 0\\
0 & 0\\
\end{array}
\right], \quad
{1 \over 2}(hh - h) = E_{22} := \left[
\begin{array}{ll}
0 & 0\\
0 & 1\\
\end{array}
\right],
$$
the latter two expressions not even being Lie brackets.  But I am not seeing how to generate $x$ or $y$ from $h$ exclusively.  What am I missing?  Sources are appreciated as well.

It's not true. Each element generates a $1$-dimensional abelian Lie algebra. However, if you read the proof, he clearly means the ideal generated by $h$, not the subalgebra. Indeed, he is trying to prove that $\mathfrak{sl}_2(\mathbb{C})$ is simple, so he wants ideals.
Also, $[h,h]=0$, not $[h,h]=I$.
