EDIT 1: All topological groups in this question are assumed to be second countable. In particular, this forces every group to be metrizable and every Lie group to have at most countably many components. In particular, all discrete groups are countable. The general case (where a Lie group is just assumed to be locally Euclidean and nothing else) could also be interesting, but for this question the second countable case is enough.
First some terminology:
A closed normal subgroup $N$ of a topological group $P$ is co-Lie in $P$, if $P/N$ is a (finite dimensional real) Lie group.
A locally compact pro-Lie group $P$ is a locally compact group such that every identity neighborhood $U\subseteq P$ contains a subgroup $N$ which is co-Lie in $P$. (This is equivalent to the fact that $P$ is the projective limit of some family of Lie groups.)
Now, there is one version of the famous Gleason-Yamabe theorem which states that
"Every locally compact group $G$ contains an open subgroup $P$ which is a pro-Lie group."
Easy examples show that such a subgroup $P$ is not unique. So, the question arises if there is a maximal one. The obvious thing to try would be to use Zorn's Lemma, but it is not clear to me if the union of a chain of open pro-Lie groups is again a pro-Lie group (it is certainly open, that is not the problem).
Any ideas on this?
EDIT 2: Note that in the case where $G$ itself is pro-Lie, it is obvious that there is a maximal open pro-Lie subgroup; just take $G$ itself. This solves the problem for the cases where
- $G$ is abelian
- $G$ is compact
- $G$ is connected (or has a finite number of components),
because in all of these cases it is well-known that $G$ is pro-Lie.
So, if there is a counter-example, it has to be a locally compact group $G$ which is non-abelian, non-compact and has infinitely many components.
Best Answer
After thinking about the problem for some time, I came up with a counter-example, so as YCor wrote in his comment, it is indeed true that there is not always a maximal open pro-Lie subgroup.
Let $A$ be a nontrivial compact group, for example $\mathbb Z/2\mathbb Z$ or the circle group $\mathbb R /\mathbb Z$. Then the countable power $A^{\mathbb N}$ is also compact. Now let $\Gamma$ be the discrete group of all permutations on $\mathbb N$ with finite support. It is clear that $\Gamma$ acts on $A^{\mathbb N}$ by permuting coordinates, so we may form the semi-direct product:
$$ G := A^{\mathbb N} \rtimes \Gamma $$
which is a locally compact group as a product of a compact and a discrete group (and tdlc if $A$ is so). It is relatively easy to see that $A^{\mathbb N}$ has no co-Lie subgroups which are $\Gamma$ invariant, so $G$ is not a pro-Lie group.
However, since $\Gamma$ is a countable union of finite permutation groups, $G$ is a countable union of compact groups which are all open in $G$. Compact groups are pro-Lie groups, so we have an increasing sequence of open pro-Lie subgroups which does not have an upper bound. So, there is no maximal open pro-Lie subgroup of $G$.