I read a comment that any two elements in $SU(2)$ each with infinite order would generate a dense subset in $SU(2)$. (Proper and dense subgroup of $\mathrm{SU}(2)$ ). The following point was made in the comments there. That:
You can also construct proper dense subgroups by taking subgroups generated by any two noncommuting infinite order elements $a,b\in SU(2)$ . (Even "most" finite order elements would work.)
First, do we know what kind of finite order elements of $SU(2)$ can generate the dense subset in $SU(2)$ ?
Second, I have tried constructing a dense subset of $SU(2)$ as follows.
Let $S=\{a,b\}$, and $G = <S>$ which is dense in $SU(2)$ according to above comments. Each word $w$ in $G$ is of the form $w=w_1 w_2 \ldots w_n $ and its reflected inverse, $ w^r = w_1^{-1},w_2^{-1} \ldots w_n^{-1}$ where $w_i$ is from $S$. Now I define a set $K = \{w^{r}w | w \in G\}$. I want to argue that $K$ is a dense subset in $SU(2)$. Here again $K$ is generated by elements in $S$ that have infinite order. If $K$ is not the dense subset of $SU(2)$, what should be added to transform it into dense subset of $G$ or $SU(2)$.
Thanks...
The argument for $SU(2)$ itself is relatively straightforward. The point is that the Lie algebra $\mathfrak{su}(2)$ has very few Lie subalgebras, which strongly constrains what the closed subgroups of $SU(2)$ look like. You can verify as an exercise that $\mathfrak{su}(2)$ has no proper nonabelian subalgebras, from which it follows that $SU(2)$ has no proper nonabelian closed connected Lie subgroups.
Now take two noncommuting elements $A, B \in SU(2)$, at least one of which has infinite order. Then they generate an infinite nonabelian subgroup of $SU(2)$, whose closure must be a closed Lie subgroup of positive dimension and hence (by the above) be all of $SU(2)$. So $A, B$ generate a dense subgroup.
For finite order elements we only need to rule out the possibility that $A, B$ generate a finite subgroup. This can be done using the classification of finite subgroups of $SU(2)$; it suffices to check that the subgroup generated by $A, B$ is not contained in a binary dihedral group (and to do this it would suffice to find two commutators which do not commute) and has order not equal to $24, 48$, or $120$.