Suppose that you have a set of matrices $S \subseteq SL(2,\mathbb{C})$, and furthermore you know that $S$ contains a continuum of matrices. (In particular, assume that there is some one-parameter family of elements $s_t \in S$, which vary continuously over some nontrivial subset of $SL(2,\mathbb{C})$ as $t$ ranges over $[0,1]$.)
Let $\langle S \rangle$ be the set of matrices generated by finite products of elements of $S$ and their inverses. Is $G$ necessarily a Lie group?
Clearly $\langle S \rangle$ is a group, and furthermore the closure of $\langle S \rangle$ in $SL(2,\mathbb{C})$ is a Lie group by the closed subgroup theorem. The question I'm asking is: if you already know there is a continuum of operations in your generating set, does this guarantee that the resulting matrix group generated is a Lie group, even if it's not closed in $SL(2,\mathbb{C})$?