Irreducibility of a 4 dimensional representation of $\mathfrak{sl}_2(\Bbb{C})$ Let $\mathfrak{g}$ be the complex Lie algebra $\mathfrak{sl}_3(\Bbb{C})$. Consider  adjoint representation $\textrm{ad} : \mathfrak{sl}_3(\Bbb{C}) \to \textrm{gl}(\mathfrak{g})$.  $\mathfrak{g}$ has the usual complex basis
$$\begin{eqnarray*} H_1 &=& \left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array}\right),
H_2 &=&   \left(\begin{array}{ccc} 0& 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{array}\right),\\
X_1 &=&  \left(\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right),
X_2 &=&   \left(\begin{array}{ccc} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right),
X_3 &=&  \left(\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{array}\right),\\
Y_1&=&  \left(\begin{array}{ccc} 0& 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right),
Y_2 &=& \left(\begin{array}{ccc} 0& 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{array}\right), Y_3 &=& \left(\begin{array}{ccc} 0& 0 & 0 \\ 0 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)\end{eqnarray*}.$$
Now by restricting $\textrm{ad}$ to just the vectors $H_1,X_1$ and $Y_1$ I can get an 8 dimensional representation of $\mathfrak{sl}_2(\Bbb{C})$. Suppose I wish to decompose this representation into irreducibles. Now I have checked that $\textrm{span}\{X_2,X_3\}$ and $\textrm{span}\{Y_2,Y_3\}$ are  irreducible 2 - dimensional subrepresentations. Now there are still the vectors $H_1,H_2,X_1,Y_1$ whose span I have tried to check is irreducible. If we write down
$$\textrm{ad}_{H_1}, \textrm{ad}_{X_1}, \textrm{ad}_{Y_1}, $$
in the basis $H_1,H_2,X_1,Y_1,X_2,X_3,Y_2,Y_3$ (in this order) we get
$$\begin{eqnarray*} \textrm{ad}_{H_1} &=&  \left(\begin{array}{cccc|cc|cc} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 2 & 0 \\ 0& 0& 0 & -2 \\ \hline &&&& -1 & 0 \\ &&&& 0 & 1 \\ \hline &&&&&&& 1 & 0 \\&&&&&&& 0 & -1  \end{array}\right)   
\textrm{ad}_{X_1} &=&  \left(\begin{array}{cccc|cc|cc} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ -2& 1 & 0 & 0 \\ 0& 0& 0 & -2 \\ \hline &&&& 0 & 0 \\ &&&& 1 & 0 \\ \hline &&&&&&& 0 & -1 \\&&&&&&& 0 & 0  \end{array}\right)   \\
\textrm{ad}_{Y_1} &=&  \left(\begin{array}{cccc|cc|cc} 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 2& -1& 0 & 0 \\ \hline &&&& 0 & 1 \\ &&&& 0 & 0 \\ \hline &&&&&&& 0 & 0 \\&&&&&&& 1 & 0  \end{array}\right).   \\
\end{eqnarray*}$$
From looking at the first $4 \times 4$ block in each matrix it seems that the span $\{H_1,H_2,X_1,Y_1\}$ is irreducible. How do I prove this formally in an elegant way without bashing?
 A: There are some typos in your formulas.
The way to decompose your representation is to first look for the eigenvalues of the Cartan subalgebra.  In the case of $\mathfrak{sl}_2$, the Cartan subalgebra is one dimensional, spanned by $H_1$.  It looks like you have eigenvalues of 2, 1, 1, 0, 0, -1, -1, and -2 (accounting for typographical error as posted).  This tells you that your representation decomposes as
$$ V_2 \oplus V_1 \oplus V_1 \oplus V_0, $$
where $V_k$ denotes the $k+1$-dimensional irreducible representation with highest weight $k$.  (That is, $V_k = \text{Sym}^k V$ where $V$ is the standard representation of $\mathfrak{sl}_2$.)
You've found the two copies of $V_1$ already.  To identify $V_2$, first find a highest weight vector $v$ of weight $2$ (i.e. an eigenvector of $\text{ad}_{H_1}$ with eigenvalue $2$) -- you've already done this in your calculation -- and then successively apply $Y_1$ to it: $Y_1(v)$ will have weight $0$ and $Y_1^2(v)$ will have weight $-2$.  Then $v, Y_1(v), Y_1^2(v)$ will be a basis of $V_2$.
I would recommend reading Chapters 11-13 in Fulton and Harris' text, as they go into these types of calculations in great depth.
