Show that $\mathfrak{sl}(2)$ is a simple Lie algebra I am trying to prove that $\mathfrak{sl}(2,\Bbb C)$ is simple.
Since this takes the $[x,y]=xy-yx$ matrix commutator bracket, this is clearly non-abelian. So to prove it is simple, we need only show that it has no trivial ideals.
Now for it to have a non trivial ideal, this ideal would need to be spanned by one or two of the basis elements of $\mathfrak{sl}(2)$, i.e. we have the basis of $\mathfrak{sl}(2)$:
$e=\begin{bmatrix}0&1\\0&0\end{bmatrix},\quad h=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\quad f=\begin{bmatrix}0&0\\1&0\end{bmatrix}$
and this non-trivial ideal must have basis:
1) $\{e,h\}$
2) $\{e,f\}$
3) $\{h,f\}$
4) $\{e\}$
5) $\{h\}$
6) $\{f\}$
But we know from the commutation relations that: $[h,e]=2e,\quad[h,f]=-2f,\quad[e,f]=h$
and this does indeed eliminate all of these basis choices, since we don't have $[\mathfrak{i},\mathfrak{sl}(2,\Bbb C)]\subseteq \mathfrak{i}$ in any case.

Am I missing anything, or making any incorrect assumptions?
Also I would be interested to know if this is the only simple complex Lie algebra of dimension $3$ up to isomorphism.
 A: It's not very hard to just do this explicitly, along the lines you started but keeping my comment in mind. Rather than $h, e, f$ I'll write $H, X, Y$. Suppose $I$ is a nonempty ideal. Then it contains some nonzero element $aH + bX + cY$. Applying $[H, -]$ to this element gives
$$b [H, X] + c [H, Y] = 2b X - 2c Y$$
from which we conclude that either $a = 0$ or $I$ contains $H$. If $I$ contains $H$, then it contains $[H, X] = 2X$ and $[H, Y] = 2Y$, and hence must be all of $\mathfrak{g} = \mathfrak{sl}_2(\mathbb{C})$. 
Otherwise, $a = 0$. Now, applying $[X, -]$ to $bX - cY$ gives
$$-c [X, Y] = -c H$$
from which we conclude that either $c = 0$ or $I$ contains $H$. If $I$ contains $H$ then as above we are done; otherwise, $I$ contains $bX$, so it contains $[X, Y] = H$, and again we're done. 
A slicker way to organize this argument is as follows. Any ideal of $\mathfrak{g}$ must in particular be invariant under the adjoint action of $H$. The adjoint action of $H$ is diagonalizable with eigenvectors $H, X, Y$, so in fact any ideal of $\mathfrak{g}$ must be a direct sum of the corresponding eigenspaces; that is, it actually turns out to be true that any ideal must have as a basis some subset of $\{ H, X, Y \}$. 
A: Concerning the last question, whether $\mathfrak{sl}(2,\mathbb{C})$ is the only complex simple Lie algebra of dimension $3$, the answer is yes. There is just one connected Dynkin diagram for rank one. Over the real numbers, there are already $2$ different simple Lie algebras of dimension $3$, see here. And over $\mathbb{Q}$ the situation is much more complicated.
