Does the group G(K) have a cocompact solvable closed subgroup? Let $K$ be a (locally compact) local field and $G$ be a linear algebraic $K$-group.
Does the topological group $G(K)$ have a cocompact solvable closed subgroup?
If $\mathrm{char}(K)=0$, it is true that $G(K)$ has a cocompact solvable closed subgroup, see Proposition 9.3 in "A. Borel and J. Tits: Groupes reductifs" (Publ. IHES, 1965: link). So the only case left is when the characteristic is positive.
It is also well-known that for any local field $K$, $\mathrm{GL}(n,K)$ has a cocompact solvable closed subgroup.
What about $G=H_{2n+1}\rtimes \mathrm{Sp}_{2n}$, which is the semidirect product of the Heisenberg group $H_{2n+1}$ with the symplectic group $\mathrm{Sp}_{2n}$?
 A: Yes. By replacing $G$ with its maximal smooth closed subgroup (whose formation commutes with any separable extension on $K$; see Lemma C.4.1 and Remark C.4.2 in the book Pseudo-reductive Groups), we may and do assume $G$ is $K$-smooth. Alternatively, if you prefer, replace $G$ with the Zariski-closure of $G(K)$ to achieve the same end. In such cases we shall describe the solvable group in rather concrete terms.
Let's first treat the case when $G$ is reductive (but possibly not connected), as inspiration for the general case.  With $G$ reductive, let $H$ be a maximal $K$-split smooth connected solvable $K$-subgroup. Explicitly, there exists a minimal parabolic $K$-subgroup $P$ of $G^0$ such that $H = S \ltimes U$ where $U = \mathscr{R}_{u,K}(P)$ is the $K$-unipotent radical of $P$ and $S$ is a maximal split $K$-torus in $P$. We claim that $G(K)/H(K)$ is compact.  
Since $H$ is smooth, the natural map $G(K)/H(K) \rightarrow (G/H)(K)$ is a homeomorphism onto an open image (via the $K$-analytic implicit function theorem and the Zariski-local structure of smooth morphisms, applied to $G \rightarrow G/H$). But this open image is full since the smooth connected $H$ is $K$-split solvable, so it is the same to show that $(G/H)(K)$ is compact.  It suffices to show that $(G^0/H)(K)=G^0(K)/H(K)$ is compact, since $G^0(K)$ is open of finite index in $G(K)$. Thus, we may and do rename $G^0$ as $G$ so that $G$ is connected.
By minimality, $P = Z_{G}(S) \ltimes U$. Note that $H$ is normal in $P$. Since $G/P$ is projective, clearly $(G/P)(K)$ is compact.  The map $G/H \rightarrow G/P$ is a right torsor for $P/H = Z_{G}(S)/S$, and $Z_{G}(S)/S$ is a $K$-anisotropic connected reductive group. Thus, $(Z_{G}(S)/S)(K)$ is compact, or in other words $(P/H)(K)$ is compact. It follows that the map $(G/H)(K) \rightarrow (G/P)(K)$ has open image over which its source is a topological torsor for the compact group $(P/H)(K)$. It is therefore enough to show that $(G/H)(K) \rightarrow (G/P)(K)$ is surjective. Even better, the Borel-Tits structure theorem for connected reductive groups over general fields gives that $G(K) \rightarrow (G/Q)(K)$ is surjective for any parabolic $K$-subgroup $Q$ of $G$ (even when ${\rm{H}}^1(K,Q)$ is nontrivial, as can happen). 

How to go beyond the reductive case? This is rather trivial when ${\rm{char}}(K)=0$, as then $G$ has a $K$-split unipotent normal smooth connected $K$-subgroup modulo which it is reductive, so let's assume ${\rm{char}}(K)=p>0$. The imperfectness of such $K$ creates many difficulties. I tried some elementary arguments involving Frobenius isogenies to try to justify direct passage to the reductive case over such $K$. Unfortunately, this involves the intervention of various non-smooth group schemes which messed up all of my attempts to try to show various subsets of topological spaces are closed. 
So in the end I failed to find an elementary reduction to the reductive case when $K$ is a local function field. All I can find is a proof that involves the full force of the structure theory of pseudo-reductive groups, which you may find somewhat heavy for the task at hand.  If someone else can find a way to rigorously justify passage to the reductive case in a more direct manner in positive characteristic then that would be swell.  In what follows, [CGP] refers to the book Pseudo-reductive Groups (2nd ed.) by Conrad, Gabber, and Prasad.

Let $H$ be a maximal $K$-split solvable smooth connected $K$-subgroup of $G$.  We claim that $G(K)/H(K) = (G/H)(K)$ is compact. Clearly $H$  contains $V := \mathscr{R}_{us,K}(G)$, the maximal $K$-split unipotent smooth connected normal $K$-subgroup of $G$, and $H/V$ is a maximal
$K$-split solvable smooth connected $K$-subgroup of $G/V$, with $G/H = (G/V)/(H/V)$. By [CGP, Cor. B.3.5], $G/V$ has $K$-unipotent radical that is $K$-wound (i.e., does not contain $\mathbf{G}_{\rm{a}}$ as a $K$-subgroup); in other words, $G/V$ is "quasi-reductive" in the sense of [CGP, Def. C.2.11]. Thus, we can replace $G$ with $G/V$ to reduce to the case that $G$ is quasi-reductive.  
Let $P$ be a minimal pseudo-parabolic $K$-subgroup of $G$ and $S \subset P$ a maximal $K$-split torus.  By [CGP, Prop. C.2.4] we have $P = Z_G(S) \cdot \mathscr{R}_{us,K}(P)$; this is a semi-direct product by the first paragraph in [CGP, Rem. C.2.33]. Let $H = S \ltimes \mathscr{R}_{us,K}(P)$, so $H$ is normal in $P$ with $H(K)$ solvable and $G(K)/H(K)=(G/H)(K)$ as in the reductive case. We claim that $G(K)/H(K)$ is compact.  As in the reductive case, $G/H \rightarrow G/P$ is a $P/H$-torsor and so $(G/H)(K) \rightarrow (G/P)(K)$ is a topological $(P/H)(K)$-torsor over its open image.  This image is full (as in the reductive case) because $G(K) \rightarrow (G/P)(K)$ is surjective [CGP, Lemma C.2.1].  Thus, as in the reductive case it suffices to prove the compactness of $(P/H)(K)$ and $(G/P)(K)$. 
We first analyze $G/P$.  Recall that $W := \mathscr{R}_{u,K}(G)$ is $K$-wound. By definition of pseudo-parabolicity we have $W \subset P$, and $G/P = (G/W)/(P/W)$ with $P/W$ a minimal pseudo-parabolic $K$-subgroup of the pseudo-reductive $G/W$ (see [CGP, Prop. 2.2.10]). Thus, to prove compactness of $(G/P)(K)$ we may and do assume $G$ is pseudo-reductive.  This case is rather subtle since $G/P$ is essentially never proper when $G$ is pseudo-reductive but non-reductive.  (The ur-example is $G={\rm{R}}_{k'/k}(G')$ for a non-separable finite extension field $k'/k$ and a nontrivial connected semisimple $k'$-group $G'$, in which case $P = {\rm{R}}_{k'/k}(P')$ for a parabolic $k'$-subgroup $P' \subset G'$. In such cases $G/P={\rm{R}}_{k'/k}(G'/P')$, and this is never proper when $P' \ne G'$ due to [CGP, Example A.5.6].) 
In view of the good behavior of pseudo-parabolic $K$-subgroups with respect to quotients of pseudo-reductive groups by central smooth connected $K$-subgroups, the general structure theorem for $G$ (in terms of the standard construction away from characteristics 2 and 3, the "generalized standard" construction in characteristics 2 and 3, and the "totally non-reduced" construction in characteristic 2: see [CGP, Prop. 10.1.4, Theorem 10.2.1(1)]) reduces the compactness of $(G/P)(K)$ to the case when $G = {\rm{R}}_{K'/K}(G')$ for a nonzero finite reduced $K$-algebra $K'$ and a smooth affine $K'$-group $G'$ whose fibers over the factor fields of $G'$ are one of the following: (i) connected semisimple, absolutely simple, and simply connected, (ii) basic exotic in characteristics 2 or 3, (iii) basic non-reduced pseudo-simple in characteristic 2. 
By [CGP, Prop. 2.2.13], the pseudo-parabolic $K$-subgroups of such a Weil restriction $G$ are precisely ${\rm{R}}_{K'/K}(Q')$ for fiberwise-pseudo-parabolic $K'$-subgroups $Q' \subset G'$. Thus, $P={\rm{R}}_{K'/K}(P')$ for a $K'$-subgroupp $P' \subset G'$ that is fiberwise minimal pseudo-parabolic.  Since $G/P = {\rm{R}}_{K'/K}(G'/P')$, so 
$(G/P)(K)=(G'/P')(K')$ as topological spaces, we can pass to the fibers over the factor fields of $K'$ separately (renaming such a factor field as $K$) so that $G$ falls into one of the cases (i), (ii), or (iii) above.  
Case (i) is easy, as then $G/P$ is projective. For case (ii), [CGP, Prop. 11.4.6(1)] identifies $G/P$ as a closed subscheme of ${\rm{R}}_{K^{1/p}/K}(G'/P')$ for a canonically determined connected semisimple (even absolutely simple and simply connected) $K^{1/p}$-group $G'$ and (minimal) parabolic $K^{1/p}$-subgroup $P'$. Thus, we get the desired compactness in these cases due to the compactness of ${\rm{R}}_{K^{1/p}/K}(G'/P')(K)=(G'/P')(K^{1/p})$ that follows from the projectivity of $G'/P'$ over $K^{1/p}$. 
How about case (iii)?  In these cases we run into a temporary snag because the natural map $\xi:G/P \rightarrow {\rm{R}}_{K^{1/2}/K}(G'/P')$ analogous to the basic exotic case turns out not to be a closed immersion, and in fact has positive-dimensional fiber over the identity point of the target; see [CGP, Rem. 11.4.5]. But fortunately in these cases a separate miracle is available: the map $\xi$ is bijective on $K$-points. To be precise, topologically this map on $K$-points is identified with $G(K)/P(K)\rightarrow G'(K^{1/2})/P'(K^{1/2})$, and this is bijective by the end of [CGP, Prop. 11.4.4]. But $G(K) \rightarrow G'(K^{1/2})$ is a homeomorphism by [CGP, Prop. 9.9.4(2)], so we win!
It remains to analyze $P/H$. Recall as in the reductive case that $P/H = Z_G(S)/S$. We claim that this group has no $K$-subgroup isomorphic to either of $\mathbf{G}_{\rm{a}}$ or $\mathbf{G}_{\rm{m}}$. The $K$-wound $W$ is a normal $K$-subgroup modulo which its quotient is $Z_{G/W}(S)/S$, so it suffices to show that $Z_{G/W}(S)/S$ has neither $\mathbf{G}_{\rm{a}}$ nor $\mathbf{G}_{\rm{m}}$ as a $K$-subgroup. But $Z_{G/W}(S)$ inherits pseudo-reductivity of $G/W$, so $Z_{G/W}(S)/S$ is a pseudo-reductive group that is $K$-anisotropic in the sense that it does not contain $\mathbf{G}_{\rm{m}}$ as a $K$-subgroup.  Hence, $Z_{G/W}(S)/S$ certainly has no proper pseudo-parabolic $K$-subgroup, so it does not contain $\mathbf{G}_{\rm{a}}$ as a $K$-subgroup either, due to [CGP, Thm. C.3.8]!
To summarize, the proof of compactness of $(P/H)(K)$ is reduced to proving the compactness of $\mathcal{G}(K)$ for any smooth connected affine $K$-group $\mathcal{G}$ that contains neither $\mathbf{G}_{\rm{a}}$ nor $\mathbf{G}_{\rm{m}}$ as a $K$-subgroup. This is Proposition A.5.7 in the paper "Finiteness theorems for algebraic groups over function fields" in Compositio 148.
A: The answer below (is a detailed edit of a previous answer which) provides a (more-or-less) self contained proof of the fact that $\mathbf{G}(K)$ contains a cocompact solvable closed subgroup.
Let $K$ be a local field and $\mathbf{G}$ a linear algebraic group defined over $K$. I denote $G=\mathbf{G}(K)$ and assume as I may that $G$ is Zariski dense in $\mathbf{G}$. Note that by [Borel's "Linear Algebraic Group, AG14.4] the Zariski closure of $G$ in $\mathbf{G}$ is itself a $K$-algebraic group whose $K$-points set is again $G$, thus replacing $\mathbf{G}$ with the Zariski closure of $G$ justifies this assumption.
Below topological notions, unless otherwise said, are taken with respect to the Hausdorff (ie $K$-analytic) topology.
Lemma 1:
If $G$ is not solvable then $\mathbf{G}$ has an irreducible $K$-representation of dimension $>1$.
Proof:
Assuming every $K$-representation is 1-dimensional, fixing an injective $K$-representation $\mathbf{G}\to \text{GL}_n$ the image of $G$ is easily seen to be in a $K$-Borel subgroup of $\text{GL}_n$ (here we use the fact that $G$ is Zariski-dense in $\mathbf{G}$).
Lemma 2:
If $G$ is not solvable then there exists a $K$-algebraic variety $\mathbf{X}$ endowed with a $K$-action of $\mathbf{G}$ and a point $x\in X=\mathbf{X}(K)$ which is not $G$-fixed such that $Gx$ is compact.
Proof:
Using lemma 1, we fix an irreducible $K$-representation $\mathbf{G}\to \text{GL}_n$ with $n>1$
and consider the associated $K$-action of $\mathbf{G}$ on $\mathbf{X}=\mathbb{P}^{n-1}$.
Observe that $X=\mathbf{X}(K)$ is compact.
By Zorn Lemma, using compactness, 
we can find a minimal closed, non-empty, $G$-invariant subset $Y\subset X$.
We fix such $Y$ and a point $x\in Y$.
By the fact that $n>1$ and the representation is $K$-irreducible we get that $x$ is not $G$-fixed.
Note that $\overline{Gx}$ is a non-empty $G$-invariant closed subspace of $Y$, thus by minimality, $Y=\overline{Gx}$.
We recall that the action of $G$ on $X$ has locally closed orbits (this is proved in the appendix of http://www.math1.tau.ac.il/~bernstei/Publication_list/publication_texts/B-Zel-RepsGL-Usp.pdf). 
It follows that $\partial(Gx)$ is a proper closed invariant subset of $Y$, hence by minimality $\partial(Gx)=\emptyset$.
We conclude that $Y=Gx$ and indeed $Gx$ is compact.
Corollary:
If $G$ is not solvable then there exists a proper $K$-algebraic group $\mathbf{H}<\mathbf{G}$ such that $H=\mathbf{H}(K)$ is cocompact in $G$.
Proof:
Fix $\mathbf{X}$ and $x\in X$ as in Lemma 2 and let $H$ be the stabilizer of $x$ in $G$ and $\mathbf{H}$ be the Zariski-closure of $H$ in $\mathbf{G}$.
By [Borel's "Linear Algebraic Group, AG14.4], $\mathbf{H}$ is a $K$-group and clearly $H=\mathbf{H}(K)$ (note however that $\mathbf{H}$ might not be the stabilizer of $x$ in $\mathbf{G}$).
We get a continuous bijective $G$-map $G/H\to Gx$. By a standard argument in the theory of locally compact groups (using Baire category) this map is open, thus $H<G$ is cocompact.
By the fact that $x$ is not $G$ fixed, $H<G$ is proper. Thus also $\mathbf{H}<\mathbf{G}$ is proper.
Theorem:
There exists a $K$-subgroup $\mathbf{S}<\mathbf{G}$ such that $S=\mathbf{S}(K)$ is solvable and cocompact in $G$.
Proof:
The collection
$$ \{\mathbf{H}\mid \mathbf{H}<\mathbf{G} \text{ is a $K$-algebraic subgrop such that $\mathbf{G}(K)/\mathbf{H}(K)$ is compact} \} $$
is not empty, as it contains $\mathbf{G}$, hence it contains a minimal element by Noetherianity. Let $\mathbf{S}$ be such a minimal element.
Denote $S=\mathbf{S}(K)$ and note that by minimality $S$ is Zariski dense in $\mathbf{S}$.
If $S$ is not solvable, then by the corollary there exists $\mathbf{H}<\mathbf{S}$ such that $H=\mathbf{H}(K)$ is cocompact in $S$, hence also in $G$,
contradicting the minimality of $\mathbf{S}$.
Thus $S$ is indeed solvable.
