Let $T$ be a torus over a $p$-adic field $F$. Let $q = f(F/\mathbb{Q}_p)$, and normalize the absolute value $| \cdot |$ on $F$ so that a uniformizer has value $\frac{1}{q}$.
Let $X(T)_F$ be the group of rational characters of $T$ which are defined over $F$. Define a homomorphism
$$H_T: T(F) \rightarrow \textrm{Hom}_{\mathbb{Z}}(X(T)_F,\mathbb{Z})$$
by $$H_T(t)(\chi) = \log_q |\chi(t)|$$
Giving the free abelian group $\textrm{Hom}_{\mathbb{Z}}(X(T)_F,\mathbb{Z})$ the discrete topology, it is easy to see that $H_T$ is continuous: the kernel is open as a finite intersection of open subgroups $\chi^{-1}\mathcal O_F^{\ast}$, as $\chi$ runs through a basis of $X(T)_F$.
It follows that $\textrm{Ker}(H_T)$ contains every compact subgroup of $T(F)$, because if $E$ is any compact subgroup of $T(F)$, then $H_T(E)$ is isomorphic to a compact, hence finite, subgroup of $\mathbb{Z}^n$, and every nontrivial subgroup of $\mathbb{Z}^n$ is infinite.
On the other hand, I have read that $\textrm{Ker}(H_T)$ itself is compact. How can one see this?