= Solution
Use the standard <Hermitian inner product> $\langle w,v\rangle=w^\dagger v$. Since $W$ is invariant under every $D_2(g)$, it is in particular invariant under
$$
D_2(g^{-1})=[D_1(g)]^\dagger.
$$
Let $v\in W^\perp$ and $w\in W$. Then
$$
\langle w,D_1(g)v\rangle
=\langle D_1(g)^\dagger w,v\rangle=0,
$$
because $D_1(g)^\dagger w\in W$ and $v$ is orthogonal to $W$. Thus $D_1(g)v$ is orthogonal to every $w\in W$, proving
$$
\boxed{D_1(g)W^\perp\subseteq W^\perp\quad(g\in G)}.
$$
Therefore $W^\perp$ is an <invariant subspace> for $D_1$. If $D_1$ is irreducible, $W^\perp$ is either $\{0\}$ or $\mathbb C^n$. In finite-dimensional <inner product space>, $(W^\perp)^\perp=W$, so these alternatives force $W=\mathbb C^n$ or $W=\{0\}$. Hence
$$
\boxed{D_1\ \text{irreducible}\ \Longrightarrow\ D_2\ \text{irreducible}}.
$$
No unitarity of $D_1$ was assumed; the <inverse-adjoint representation> is exactly what makes this <orthogonal complement> argument work.
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