First, it should be noted that something much more general is true, and documented with proof in the literature. Suppose $F$ is the fraction field of a rank-1 henselian valuation ring $R$ (for simplicity feel free to assume $R$ is a complete discrete valuation ring, though one only needs that $R$ is henselian rather than complete, which in the rank-1 case is equivalent to the associated absolute value on $F$ admitting a unique extension to every finite extension of $F$: see 2.3.1 and 2.4.3 in Berkovich's paper in IHES 78 for that equivalence). For any affine $F$-scheme $X$ of finite type, it is well-posed to say $X(F)$ is *bounded* (with respect to $X$) if some some closed immersion $j:X \hookrightarrow \mathbf{A}_F^n$ the closed subset $X(F) \subset F^n$ is bounded; this is independent of the choice of $j$ and if $F$ is locally compact then it is equivalent to compactness of $X(F)$. If $G$ is a connected reductive $F$-group such that $G(F)$ is bounded then obviously $G$ is $F$-anisotropic (i.e., $G$ cannot contain ${\rm{GL}}_1$ as an $F$-subgroup, since boundedness is inherited by closed subschemes and $F^{\times}$ is not bounded (with respect to ${\rm{GL}}_1$). The great fact is that the converse is true: this is due to Rousseau, and a slick proof is given by Gopal Prasad in his paper *Elementary proof of a theorem of $\dots$* in Bulletin de la SMF, tome 110 (1982), pp. 197-202.

My recollection is that Prasad's proof treats all such $G$ on equal footing, not requiring preliminary treatment of the case of tori. You might nonetheless consider this to be rather heavy for just treating tori, since Prasad's argument involves a lot of the structure theory of reductive groups (which admittedly simplifies to a triviality for tori).

Here is an argument that is a variant on Venkataramana's answer (comes down to the same fact about integral units being "bounded", but doesn't invoke facts about the semisimplicity of the algebraic representation theory of general tori over fields and includes some details that might make the argument look more complicated than it really is). Over any field $k$, we have an anti-equivalence of categories between $k$-tori $T$ and discrete ${\rm{Gal}}(k_s/k)$-modules $L$ that are finite free as $\mathbf{Z}$-modules (assigning to $T$ the geometric character lattice ${\rm{X}}(T) := {\rm{Hom}}_{k_s}(T_{k_s}, {\rm{GL}}_1)$ equipped with its natural (discrete) Galois action).

Let $\Gamma = {\rm{Gal}}(k'/k)$ for a finite Galois extension $k'/k$ that splits $T$. Then $L := {\rm{X}}(T)$ is a quotient of a finite direct sum of copies of $\mathbf{Z}[\Gamma]$ as a discrete $\Gamma$-lattice. Rationalizing to make the $\Gamma$-action semisimple, the $k$-anisotropicity of $T$ is *exactly* the condition that
${\rm{X}}(T)_{\mathbf{Q}}$ has no copy of the trivial representation. Equivalently, each map $\mathbf{Q}[\Gamma] \rightarrow {\rm{X}}(T)_{\mathbf{Q}}$ as $\Gamma$-lattices kills the trivial representation $\mathbf{Q}$. Hence, the $\Gamma$-lattice ${\rm{X}}(T)$ is a quotient of a direct sum of copies of $\mathbf{Z}[\Gamma]/\mathbf{Z}$. This latter Galois lattice corresponds to the $k$-subgroup $U^1$ of norm-1 units in ${\rm{R}}_{k'/k}({\rm{GL}}_1)$.

Going from $\Gamma$-lattices back to $k$-tori split by $k'$ turns this quotient presentation into a *closed immersion* of $k$-groups $T \hookrightarrow \prod_{j=1}^n U^1$, so it suffices to show that if $k$ is a rank-1 henselian valued field then $U^1(k)$ is bounded with respect to $U^1$. But with some review of the construction of Weil restriction through a finite extension of fields, one can show that if $k'/k$ is a finite extension of fields and $X'$ is an affine $k'$-scheme of finite type then the canonical bijection ${\rm{R}}_{k'/k}(X')(k) = X'(k')$ is a topological isomorphism, and that a closed subset if bounded with respect to ${\rm{R}}_{k'/k}(X')$ if and only if it is bounded with respect to $X'$.

In this way we are reduced to showing that the kernel of ${\rm{N}}_{k'/k}: {k'}^{\times} \rightarrow k^{\times}$ is bounded inside the "hyperbola" $uv=1$ within $k' \times k'$. But that kernel is a closed subgroup of $O_{k'}^{\times}$, so the desired boundedness is clear.

QED

cocharacter. Notice the 'rationality' in that description; it means that we have a map $F^\times \to S(F)$, not just $E^\times \to S(E)$. (I am using that the lattices of rational characters and of rational cocharacters are dual, so that one has some non-0 of the former if and only if one has some non-0 of the latter.) $\endgroup$ – LSpice Feb 25 '17 at 4:36closed$F$-subgroup of $S$ (much better than just a nontrivial $F$-homomorphism from ${\rm{GL}}_1$), so $F^{\times}$ with its natural analytic topology is a closed subgroup of $S(F)$, ruling out compactness of the latter. $\endgroup$ – nfdc23 Feb 25 '17 at 5:045more comments