Let $\frak{h} \subseteq \frak{so}(4,\Bbb{C})$ be the subalgebra consisting of matrices of the form $$\begin{pmatrix} 0 & a & 0 & 0 \\ -a & 0 & 0 & 0 \\ 0 & 0 & 0 & b \\ 0 & 0 & -b & 0 \end{pmatrix}.$$ Show that $\frak{h}$ is a Cartan subalgebra and find the corresponding root decomposition.
For now I'm just trying to show that $\frak{h}$ so defined is a Cartan subalgebra. Clearly it is commutative; i.e., clearly $[x,y] = 0$ for all $x,y \in \frak{h}$ which is equivalent to $xy=yx$ for all $x,y \in \frak{h}$, because the Lie bracket is just the matrix commutator.
Now I am trying to show that every element is semisimple; that is, the adjoint map $\text{ad } x : \frak{so}(4,\Bbb{C}) \to \frak{so}(4,\Bbb{C})$ is a diagonalizable operator for every $x \in \frak{h}$. One way of showing that this operator is diagonalizable is to show that the sum of the dimension of the eigenspaces is the same as $\dim \frak{so}(4,\Bbb{C})$. But that seems a little difficult. What is $\dim \frak{so}(4,\Bbb{C})$? How does one describe the elements of $\frak{so}(4,\Bbb{C})$; does it have a nice basis?
I haven't really thought about showing the centralizer of $\frak{h}$ equals $\frak{h}$, because I don't really know how to describe the elements of $\frak{so}(4,\Bbb{C})$.
Let $x=x_{a,b}$ be the matrix $$\begin{pmatrix}0&a&0&0\\-a&0&0&0\\0&0&0&b\\0&0&-b&0\end{pmatrix}.$$ Treat $x$ as an element of $\mathfrak{gl}(4,\mathbb{C})=\operatorname{End}_{\Bbb C}(\mathbb{C}^4)$. Then notice that $x$ is diagonalizable (as a linear operator), and so it is a semisimple linear operator. Use this to show that $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ is a semisimple linear operator on $\mathfrak{gl}(4,\mathbb{C})$. However, $\mathfrak{so}(4,\mathbb{C})$ is an invariant subspace of $\mathfrak{gl}(4,\mathbb{C})$ under $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ (as $x\in\mathfrak{so}(4,\mathbb{C})$ and $\mathfrak{so}(4,\mathbb{C})$ is a Lie subalgebra of $\mathfrak{gl}(4,\mathbb{C})$), so $$\operatorname{ad}_{\mathfrak{so}(4,\mathbb{C})}x=\left.\left(\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x\right)\right|_{\mathfrak{so}(4,\mathbb{C})}.$$ This shows that $\operatorname{ad}_{\mathfrak{so}(4,\mathbb{C})}x$ is a semisimple linear operator on $\mathfrak{so}(4,\mathbb{C})$.
In fact, you should be able to see that the eigenvalues of $x$ are $\pm ai$ and $\pm bi$. Therefore, the eigenvalues of $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ are
Intersecting the eigenspaces of $\operatorname{ad}_{\mathfrak{gl}(4,\mathbb{C})}x$ with $\mathfrak{so}(4,\mathbb{C})$, we see that the $6$-dimensional subspace $\mathfrak{so}(4,\mathbb{C})$ of $\mathfrak{gl}(4,\mathbb{C})$ consists of the following eigenspaces
The $2$-dimensional eigenspace corresponding to the eigenvalue $0$ is precisely $\mathfrak{h}$ itself. This implies that the centralizer of $\mathfrak{h}$ is $\mathfrak{h}$ itself. (Technically speaking, you are proving that $\mathfrak{h}$ is a maximal toral subalgebra of $\mathfrak{so}(4,\mathbb{C})$. However, for finite-dimensional semisimple Lie algebras, Cartan subalgebras are identical to maximal toral subalgebras.)
For simplicity, write $\mathfrak{g}=\mathfrak{so}(4,\mathbb{C})$. Let $\alpha$ and $\beta$ be the linear functionals in $\mathfrak{h}^*=\operatorname{Hom}_{\Bbb C}(\mathfrak{h},\Bbb C)$ such that $\alpha(x_{a,b})=ai$ and $\beta(x_{a,b})=bi$. For $\gamma\in\mathfrak{h}^*$, define $\mathfrak{g}_{\gamma}$ to be the intersection of the eigenspaces of $x_{a,b}$ corresponding to the eigenvalue $\gamma(x_{a,b})$. Then, $$\mathfrak{g}=\mathfrak{h}\oplus \bigoplus_{s,t=\pm1}\mathfrak{g}_{s\alpha+t\beta}$$ with $\mathfrak{h}=\mathfrak{g}_0$. Note that each $\mathfrak{g}_{s\alpha+t\beta}$ for $s,t=\pm 1$ is a $1$-dimensional subspace of $\mathfrak{g}$.