Let $X$ be a smooth geometrically integral variety over $\mathbf{Q}$, having a $\mathbf{Q}$-point.
We say that $X$ has *the real approximation property* if $X(\mathbf{Q})$ is dense in $X(\mathbf{R})$.

Let now $X$ be a homogeneous space of a connected linear algebraic group $G$ defined over $\mathbf{Q}$. Since $X$ has a $\mathbf{Q}$-point, we may write $X=G/H$, where $H$ is a $\mathbf{Q}$-subgroup of $G$, not necessarily connected.

Question 1.Does real approximation hold for any homogeneous space $X=G/H$ of any connected linear algebraic $\mathbf{Q}$-group $G$, where $H$ is not necessarily connected?

Serre proved real approximation for algebraic tori over $\mathbf{Q}$ (i.e. when $G$ is a torus and $H=1$).
Sansuc proved real approximation for connected linear algebraic groups over $\mathbf{Q}$
(i.e. when $G$ is any connected linear algebraic group over $\mathbf{Q}$ and $H=1$).
One can prove real approximation for homogeneous spaces with *connected stabilizer*
(i.e. when $G$ is a connected linear algebraic group over $\mathbf{Q}$, and $H\subset G$ is any connected $\mathbf{Q}$-subgroup).
In the general case (i.e. when $H$ is non-connected) I expect the negative answer, but I cannot construct a counter-example.

The question can be stated in terms of Galois cohomology. Assume for simplicity that $G=\mathrm{SL}_n$, then $H^1(\mathbf{Q},G)=1$ and $H^1(\mathbf{R},G)=1$. Any orbit of $G(\mathbf{R})$ in $X(\mathbf{R})$ is open. If such an orbit contains a $\mathbf{Q}$-point, then $\mathbf{Q}$-points contained in this orbit are dense in this orbit, because our $G=\mathrm{SL}_n$ has real approximation. We denote by $G(\mathbf{R})\backslash X(\mathbf{R})$ the set of orbits of $G(\mathbf{R})$ in $X(\mathbf{R})$. Now our question is, whether the canonical map $G(\mathbf{Q})\backslash X(\mathbf{Q})\to G(\mathbf{R})\backslash X(\mathbf{R})$ is surjective. There is a canonical bijection $$ G(\mathbf{Q})\backslash X(\mathbf{Q})=\mathrm{ker}[H^1(\mathbf{Q},H)\to H^1(\mathbf{Q},G)]=H^1(\mathbf{Q},H) $$ and similarly $G(\mathbf{R})\backslash X(\mathbf{R})=H^1(\mathbf{R},H)$, so in this case Question 1 can be stated as follows:

Question 2.Is it true that for any linear algebraic group $H$, defined over $\mathbf{Q}$ (not necessarily connected or abelian) the localization map $\mathrm{loc}_\infty\colon H^1(\mathbf{Q},H)\to H^1(\mathbf{R},H)$ is surjective?

If the group $H$ is connected or abelian, then the localization map $\mathrm{loc}_\infty$ is surjective
(Sansuc's proof in the case of finite abelian $H$ uses the Chebotarev density theorem
and the fact that a decomposition group of $\infty$ is isomorphic to $\mathbf{Z}/2\mathbf{Z}$, hence cyclic).
In the general case I expect the negative answer,
i.e. that there exists a *finite* non-abelian algebraic group $H$
over $\mathbf{Q}$ for which the localization map
$\mathrm{loc}_\infty$ is *not surjective*, but I cannot construct such a counter-example.