= Solution
Let $V\ne0$ be finite-dimensional. Choose an $H$-<eigenvector> and apply $e$ repeatedly until the last nonzero vector $v$ in that chain. Such a last vector exists because $e$ raises the $H$-<eigenvalue> by two and $H$ has only finitely many eigenvalues. Then $ev=0$ and $Hv=lv$ for some $l$.
The nonzero $f$-chain cannot continue indefinitely, since its weights are distinct. Part (a), with its zero endpoint included, gives $l\in\mathbb Z_{\ge0}$ and the <basis> $v,fv,\ldots,f^lv$. The raising formula in part (a) shows its span is a <submodule> isomorphic to $A_l$.
To prove $A_l$ <irreducible>, a nonzero <submodule> stable under diagonal $H$ contains a <weight vector>: <polynomial interpolation> in the distinct eigenvalues projects onto individual <representation weight> lines. Repeated raising reaches $v_0$, and repeated lowering then reaches every $v_k$. Thus the <submodule> is all of $A_l$. Distinct $l$ give different dimensions $l+1$, so the modules are nonisomorphic. Every finite-dimensional <irreducible> <module> contains one of these nonzero simple submodules and hence equals it. Therefore
$$
\boxed{\{A_l:l\ge0\}\text{ is the complete list of finite-dimensional simple modules}.}
$$
The zero <module> has no nonzero simple <submodule> and is dealt with as the empty <direct sum> in part (d).
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