Say I have a Hermitian operator, $H^{\dagger} = H$.
Does there always exist at least one operator that may or may not be Hermitian, $O^{\dagger} \neq O$, such that:
$$ H = OO^{\dagger} $$
It makes sense to me that this should be the case, since I can't imagine what a counterexample would look like, but also it would be nice to know that a proof exists.
The operators of the form $OO^\dagger$ are precisely the Hermitian operators with non-negative spectrum (often called "positive operators").
An easy counterexample in $M_2(\mathbb C)$ would be $$H=\begin{bmatrix} 1&0\\0&-1\end{bmatrix},$$ with eigenvalues $1,-1$.