Edit: I'm specializing this question to the compact case I'll ask about the noncompact case as a new question.
Let $ G $ be a compact connected semisimple Lie group.
Do there always exist two finite order elements of $ G $ which generate a dense subgroup?
Example:
$$
\frac{i}{\sqrt{2}}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}
$$
and
$$
\begin{bmatrix} \overline{\zeta_{16}} & 0 \\ 0 & \zeta_{16} \end{bmatrix}
$$
generate a dense subgroup of $ SU_2 $.
This question is partially inspired by
Generating finite simple groups with $2$ elements
which shows that every finite simple group is 2-generated. Indeed even every finite quasisimple group is 2-generated Is every finite quasi-simple group generated by 2 elements? (however this fails for infinite simple groups:
$ \mathrm{PSL}_2(\mathbb{Q}) $ is not 2-generated indeed not even finitely generated).
This is a cross-post of https://math.stackexchange.com/questions/4537024/dense-subgroups-generated-by-two-finite-order-elements
an answer posted there points out that every compact semisimple Lie group can be topologically generated by 2 infinite order elements.
Best Answer
You might want to look at MR0034766 (Kuranishi 1949, link). It deals with the connected semisimple case, which might be what you're after.
There's also a 1999 Proc AMS paper by M. Field, (freely accessible at AMS site — MR number for subscribers: MR1618662), which shows that the set of pairs generating a dense open subset of $G$ is non-empty Zariski open in $G \times G$ when $G$ is compact connected semisimple. This should lead to a proof that the elements can be taken to have finite order in this case, as the set of pairs of elements of finite order is dense in $G\times G$. Of course, in the non-compact case some extra condition on $G$ is required.