Surely it's easier to check whether $H'$ is simply-connected by inspecting the co-root lattice...? For the example of $G_2$ containing $SO(4)$ that Allen mentions, we have a pseudo-Levi subalgebra of type $A_1\times \tilde{A}_1$ where $\{(3\alpha+2\beta),\alpha\}$ is a basis of simple roots. Now the cocharacter $(3\alpha+2\beta)^\vee=\alpha^\vee+2\beta^\vee$, so the lattice:
${\mathbb Z}(3\alpha+2\beta)^\vee+{\mathbb Z}\alpha^\vee = {\mathbb Z}\alpha^\vee+{\mathbb Z}(2\beta^\vee)$
is of index two in the cocharacter lattice ${\mathbb Z}\alpha^\vee+{\mathbb Z}\beta^\vee$ for $T$.
In fact this allows you to determine exactly what the pseudo-Levi subgroup is in each case.
For the maximal pseudo-Levis there's an easier trick to find non-simply-connected ones: if $s\in T$ and $L=Z_G(s)$ then $Z(L)/Z(L)^\circ$ is generated by $s$, by a result of Eric Sommers. So we can see almost immediately that hardly any maximal pseudo-Levi subgroups are simply-connected. For example, the pseudo-Levi of $F_4$ which is of type $C_3\times A_1$ has a cyclic centre, so it can't be isomorphic to $Sp_6\times SL_2$. Specifically, it is isomorphic to $(Sp_6\times SL_2)/\{ \pm (I,I)\}$.
EDIT: A mistake with this is that Sommers' result only holds for adjoint type groups. More generally we have $Z(L)/(Z(L)^\circ Z(G))$ is generated by $s$. Of course this makes no difference for type $F_4$.
A natural condition is that $G$ has finitely many connected components.
One can easily reduce this case to the connected group case, and then to the compact group case, as all maximal compact subgroups in a connected Lie group are conjugated.
Then the representation variety $\text{Hom}(\Gamma,G)$ is compact and local rigidity, that is vanishing of $H^1(\Gamma,\mathfrak{g})$, guarantees the finiteness of the number of $G$-orbits. Here $\mathfrak{g}$ denotes the Lie algebra of $G$. The fact that $H^1$ vanishes could be deduced from the fact that every isometric action of $\Gamma$ on $\mathfrak{g}$ has a fixed point, by averaging an orbit.
Best Answer
7 years ago I wrote in an unfinished paper a proof of this result; I didn't know then if it was original but it indeed seems to be known to a number of specialists although I'm not aware of a written complete proof. Since I don't want anymore to complete this paper, I reproduce the proof here (not self-contained, relying on [LW] below) as Proposition 5; possibly Uri has a more direct approach. I include closely related Proposition 1, which was my original goal.
Proposition 1 Let $G$ be a compact group (not necessarily Lie). Consider the action by $G$ by conjugation on $\mathcal{S}(G)$ (the space of subgroups of $G$, endowed with the Chabauty topology). Then the orbits of $G_0$ (connected component of 1) coincide with the connected components of $\mathcal{S}(G)$.
As a corollary, for $G$ a compact group, $\mathcal{S}(G)$ is totally disconnected if and only if $G_0$ is central in $G$ (equivalently, $G/Z(G)$ is totally disconnected), a result obtained in [FG]. More generally, it can be checked that for an arbitrary locally compact group $G$, $\mathcal{S}(G)$ is totally disconnected if and only if either totally disconnected, or $G$ is pointwise elliptic with $G_0$ central. ($G$ pointwise elliptic means that every single element generates a cyclic subgroup with compact closure.)
Lemma 2 Let $G$ be a compact Lie group. Let $F$ be a finite group. Then there are finitely many conjugacy classes of homomorphisms $F\to G$.
Proof. The space $\mathrm{Hom}(F,G)$ can be described as a real algebraic variety, so has finitely many components. By [LW], all homomorphisms in a given component are conjugate under $G_0$. \end{proof}
Lemma 3 Let $G$ be a compact Lie group. Let $S$ be a connected compact semisimple Lie group. Then there are finitely many conjugacy classes of homomorphisms $S\to G$.
Proof. The space $\mathrm{Hom}(S,G)$ is canonically identified with the space of Lie algebra homomorphisms $\mathfrak{s}\to\mathfrak{g}$, which is an algebraic variety as well, so has finitely many connected components. So we can again use [LW].
Lemma 4 Let $G$ be a compact Lie group. Let $D$ be a connected compact abelian Lie group. Then there are countably many conjugacy classes of homomorphisms $D\to G$.
Proof. Any such homomorphism maps into a maximal torus, and all maximal tori are conjugate [Bourbaki, Groupes et Algèbres de Lie, Chap 9, §2.2]. This reduces to the case when $G$ is a torus, but then the result just follows from the fact that $\mathrm{Hom}((\mathbf{R}/\mathbf{Z})^a,(\mathbf{R}/\mathbf{Z})^b)\simeq\mathbf{Z}^{ab}$, which is countable.
Lemma 5 Let $G$ be a compact Lie group. Then there are countably many conjugacy classes of closed connected subgroups in $G$.
Proof. Since there are finitely many isomorphism types of compact semisimple groups in a given dimension, we can fix this type, so by Lemma 3, it is enough to consider the set of closed connected subgroups whose semisimple part is a given subgroup $S$. Therefore, it is enough to count conjugacy classes of closed abelian connected subgroups in $N(S)/S$, where $N(\cdot)$ denotes the normalizer, and Lemma 4 applies.
Proposition 6 Let $G$ be a compact Lie group. Then there are countably many conjugacy classes of closed subgroups in $G$.
Proof. By Lemma 5, we can consider those subgroups for which $H_0=L$ is given. So it is enough to count conjugacy classes of finite subgroups in $N(L)/L$, there are countably many because of Lemma 2.
Proof of Proposition 1 We begin with the case when $G$ is a Lie group. Since $G_0$ has finite index in $G$, Proposition 6 implies that there are countably many $G_0$-conjugacy classes of closed subgroups in $G$. Let $C$ be a connected component of $\mathcal{S}(G)$. Clearly, it is invariant under conjugation by $G_0$. We just obtained that the quotient of $C$ by the action of $G_0$ is countable. It is also a compact connected Hausdorff space, so is reduced to a point (it is Hausdorff because the acting group is compact, see [Bourbaki, Topologie Générale, III, §4.1]. This means that $G_0$ acts transitively on each connected component of $\mathcal{S}(G)$, and we are done.
Now let us deal with a general compact group $G$. (It is no longer true in general that the set of connected components of $\mathcal{S}(G)$ is countable). We write $G$ as a projective limit of compact Lie groups $G/K_i$. Clearly, if two elements $H,H'$ of $\mathcal{S}(G)$ are conjugate under $G_0$, then they lie in the same component. Conversely, suppose they belong to the same connected component of $\mathcal{S}(G)$. Then since $K_i$ is compact, the projection map $\mathcal{S}(G)\to\mathcal{S}(G/K_i)$ is continuous. Because of the case of Lie groups, there exists $g_i\in G_0$ such that $(g_iHg_i^{-1}K_i=H'K_i)$ (we use the fact that the inverse image of the unit component of $G/K_i$ is the unit component of $G$). Let $g$ be a limit point of $(g_i)$. If $h\in H$, then $g_ihg_i^{-1}$ can be written as $h'_ik_i$ with $h'_i\in H'$ and $k_i\in K_i$. Necessarily $k_i\to 1$, so $h'_i=g_ihg_i^{-1}k_i^{-1}$ also converges, necessarily to an element $h'$ of $H'$. We have $ghg^{-1}=h'$. This proves that $gHg^{-1}\subset H'$, and the converse inclusion $g^{-1}H'g\subset H'$ is strictly similar.
[FG] S. Fisher, P. Gartside. On the space of subgroups of a compact group II. Topology Appl. 156 (2009) 855-861
[LW] D.H. Lee, T.S. Wu. On conjugacy of homomorphisms of topological groups. Illinois J. Math. 13 1969 694-699
in [LW] I refer to Theorem 2.6: if $G,H$ are compact Lie groups, then the connected components of $\mathrm{Hom}(G,H)$ are precisely the $H_0$-orbits (for action by post-conjugation).