Let $k$ be a local field of characteristic zero, and $G$ a linear algebraic group over $k$. Let $X$ be a variety over $k$, and $G \times X \rightarrow X$ a group action defined over $k$. When we pass to the $k$-rational points, we get a continuous action of $G(k)$ on $X(k)$.
1 . Are the $G(k)$-orbits always closed in $X(k)$?
2 . If $x_0 \in X(k)$, is the natural continuous map $G(k)/\textrm{Stab}(x_0) \rightarrow G(k).x_0$ a homeomorphism?
For the first question, the answer is in general no. For example, $\textrm{GL}_n(\mathbb{R})$ acts continuously on $\mathbb{R}^n$, and there are two orbits, the zero vector and its complement.
However, if $G$ is a topological group acting continuously on a topological space $X$, then the quotient sapce $G \setminus X$ is Hausdorff if and only if all orbits are closed. This is because a point in $G \setminus X$ is closed if and only if its preimage is closed in $X$.
For the second question, the answer is yes. Since $k$ is locally compact Hausdorff, so are $G(k)$ and $X(k)$, and $G(k)$ has a countable basis. So there is a more general result we can apply. This is from J.S. Milne's notes on modular forms.
Lemma 1: If $G$ is a nonempty locally compact Hausdorff topological space, and $G$ is the countable union of closed sets $V_n$, then at least one of the $V_n$ contains a nonempty open set.
Proof: Suppose no $V_n$ contains a nonempty open set. Let $U_1$ be a nonempty open set in $G$ whose closure is compact. Then $U_1 \not\subseteq V_1$, which means $U_1 \cap V_1 \subsetneq U_1$. Now $U_1 \cap V_1$ is closed in $U_1$, so we can find a nonempty open set $U_2 \subseteq U_1$ whose closure $\overline{U_2}$ is contained in $U_1$ and disjoint from $U_1 \cap V_1$. This is because compact sets form a neighborhood basis of any point.
Now $U_2 \not\subseteq V_2$, so $U_2 \cap V_2 \subsetneq V_3$, and we can find an open $U_3$ whose closure $\overline{U_3}$ is compact and disjoint from $U_2 \cap V_2$.
Iterating, we have open sets $U_1 \supseteq U_2 \supseteq \cdots$ with compact closures $\overline{U_1} \supseteq \overline{U_2} \supseteq \cdots$ such that $\overline{U_i}$ is disjoint from $U_{i-1} \cap V_{i-1}$ for $i \geq 2$. The intersection of the compact sets $\overline{U_i} : i = 2, 3, ...$ must be nonempty. Let $p$ be in this intersection. For each $i \geq 1$, we have $p \in U_i$. And for each $i \geq 2$, we have $p \not\in U_{i-1} \cap V_{i-1}$. Hence $p$ is not in any of the $V_i$, which is impossible. $\blacksquare$
Lemma 2: Let $G$ be a locally compact Hausdorff topological group with a countable basis, acting continuously on a locally compact Hausdorff space $X$. If $x_0 \in X$, then the natural map $G/ \textrm{Stab}(x_0) \rightarrow G.x_0$ is a homeomorphism.
Proof: We may replace $X$ by the orbit of $x_0$ and assume $G$ acts transitively on $X$.
Let $H$ be the stabilizer of $x_0$. The universal property of quotients gives us a continuous bijection $G/H \rightarrow X$, which we need to show is open.
Let $U$ be open in $G$. Let $gx_0$ be a point of $Ux_0$. We need to find an open neighborhood $W$ of $gx_0$ which is contained in $Ux_0$. Of course, we can choose $g$ to be in $U$.
Standard topological group shenanigans gives us a neighborhood $V$ of $1_G$ such that $V = V^{-1}$, and $gV^2 \subseteq U$. Furthermore, we can choose $V$ to be compact. Since $G$ has a countable basis, it is easy to see that $G$ is the union of countably many translates $g_nV$ for some $g_n \in G$.
Each $g_nV$ is compact, hence so is the image $g_nVx_0$. So $X$ is the countable union of the closed sets $g_nVx_0$. By the first lemma, one of the closed sets $g_nVx_0$ contains a nonempty open set. These closed sets are translated around homeomorphically by $G$. So this means that $Vx_0$ must contain a nonempty open set $W$. Let $hx_0 \in W$ for some $h \in V$. Then $gx_0 = gh^{-1}hx_0$ lies in the open set $gh^{-1}W$. But since $V^2 \subseteq U$, we have $$gh^{-1}W \subseteq gV^2x_0 \subseteq gVx_0 \subseteq Vx_0 \subseteq Ux_0$$