Original answer
This isn't an answer, but since there has been little activity on this (very interesting) question I guess I might as well say it.
What if we consider a set of generators of $G_1$ and keep extending it by adding some elements so that it generates $G_n$ at each step. Then, looking at the Cayley graphs we can put some distance on each group, making this a sequence of finite metric spaces. The idea would be to do this in such a way so that the spaces are converging in the Gromov-Hausdorff sense to some metric space $(X,d)$ and the uniform probabilities $\mu_n$ on each $(G_n,d_n)$ are converging to some probability measure $\mu$ on $X$.
Added later...
Compactification and weak limit of probabilities in the case of finite Abelian cyclic groups
Consider the case when all the groups are finite cyclic so that $G_k = \mathbb{Z}_{n_k}$ (i.e. the finite cyclic group of $n_k$ elements) for some non-decreasing sequence of natural numbers each of which divides the next $n_1 | n_2 | \cdots$.
Let $S = \mathbb{R}/\mathbb{Z}$, we can identify each $G_k$ with the subgroup $(\frac{1}{n_k}\mathbb{Z})/\mathbb{Z} \subset S$. The uniform probability measures on these finite subgroups either converge weakly to Lebesgue measure on $S$ or (if the sequence of numbers $n_k$ is eventually constant) are eventually equal to a constant measure supported on a finite subgroup.
Compactification and weak limit of probabilities in the case of general Abelian finite groups
Consider now the case in which all groups $G_k$ are abelian. By the clasification of finite abelian groups, one can decompose each $G_k$ into a direct sum of cyclic groups with orders that are powers of primes. This implies that one can obtain a group isomorphic to $G_{k+1}$ from $G_k$ by replacing some of these powers of primes by higher powers, and by forming the direct product with another cyclic group of power of prime order.
Let $G = S^{\mathbb{N}}$ be the cartesian product of countably many copies of $S$. This is a compact and metrizable group. One can identify each group $G_k$ with a subgroup of $G$ generated by a finite number of elements with power of prime order and only one non-null corrdinate. The extension from $G_k$ to $G_{k+1}$ is obtained by replacing one of these generators by another with the same non-null coordinate but whose order is a higher power of the same prime number, or by adding a new generator with power of prime order whose only non-null coordinate is distinct from that of all the other generators.
This procedure gives rise to an increasing family of subgroups of $G$. The sequence of uniform measures $\mu_n$ on these groups can be seen to have a weak limit since each projection to a coordinate does (it is either eventualy a constant uniform measure on a finite subgroup of prime power order, or converges to Lebesgue measure on the circle).
Further directions
The answer to the question posed here implies that there are countable groups that are not a subgroup of a compact group. Hence it might be possible to construct a counter-example using one of these countable groups as the union of the $G_n$. The simplest possible candidate seems to be obtained by taking $G_k = \text{SL}(2,\mathbb{F}_{2^k})$ where $\mathbb{F}_p$ is the field with $p$ elements.
Also, here several countable groups which contain all finite subgroups are defined. This might serve to reduce the discussion to one concrete chain such as $G_1 = S_3, G_{n+1} = S_{G_n}$ where $S_G$ denotes the group of permutations of the elements of $G$ (which contains $G$ a subgroup since each element of $G$ acts on $G$ as a permutation).