Let $\mathfrak{g}$ be a Lie algebra, $\mathfrak{a}$ a non-zero abelian ideal in $\mathfrak{g}$, and $\mathfrak{s}$ a vector space such that $$\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{s}.$$
Let $X \in \mathfrak{a}$ and $Y \in \mathfrak{g}$. Then, $$[X, \mathfrak{a}] = 0 \tag{1}$$ $$[X, \mathfrak{s}] \subset \mathfrak{a} \tag{2}$$ $$[Y, \mathfrak{a}] \subset \mathfrak{a} \tag{3}$$ I know that (1) follows because $\mathfrak{a}$ is abelian and (3) is because $\mathfrak{a}$ is assumed to be an ideal. I think (2) also follows from $\mathfrak{a}$ being an ideal, because $\mathfrak{a}$ being an ideal implies $$[\mathfrak{s}, X] \subset \mathfrak{a}$$ but since $\mathfrak{a}$ is a subspace, $-[\mathfrak{s}, X] = [X, \mathfrak{s}]$ is also in $\mathfrak{a}$. Is this right?
I am following a proof of Cartan's criterion for semisimplicity, which says that because of (1)-(3), under the splitting $\mathfrak{g} = \mathfrak{a} \oplus \mathfrak{s}$ we have $$\text{ad}_X = \begin{pmatrix} 0 & * \\ 0 & 0 \end{pmatrix} \\ \text{ad}_Y = \begin{pmatrix} * & * \\ 0 & * \end{pmatrix}.$$
How does (1)-(3) imply that $\text{ad}_X$ and $\text{ad}_Y$ must be of the above form? More specifically, why are they necessarily $2\times 2$ and why are the nonzero elements in each matrix where they are?
These are not $2\times2$ matrices. They are block matrices. Let $\{a_1,\ldots,a_n\}$ be a basis of $\mathfrak a$ and let $\{s_1,\ldots,s_m\}$ be a basis of $\mathfrak s$. Then $\operatorname{ad}_X(a_j)=0$ for each $j\in\{1,2,\ldots,n\}$ and $\operatorname{ad}_X(s_j)$ is a linear combination of $a_i$'s, for each $j\in\{1,2,\ldots,m\}$. Therefore, the first $n$ columns of the matrix of $\operatorname{ad}_X$ with respect to the basis $\{a_1,\ldots,a_n,s_1,\ldots,s_m\}$ are null columns and the final $m$ entries of the remaining columns are all equal to $0$. So, that matrix is a block matrix of the form $\left[\begin{smallmatrix}0&*\\0&0\end{smallmatrix}\right]$. A similar argument applies to $\operatorname{ad}_Y$.