The answer is No! That is:
Let $G$ be a locally compact group. Then $G$ has a compact open subgroup or a discrete infinite cyclic subgroup.
Let $C$ be the connected component of $G$, and recall the following facts (see [1], [2] and references therein):
(1) If $G$ is totally disconnected, then $G$ contains a compact open subgroup.
(2) $C$ is a normal subgroup of $G$, and $G/C$ is totally disconnected.
(3) If $G$ is connected, then every neighborhood of 1 contains a compact normal subgroup $K$ such that $G/K$ is a connected Lie group.
Assume $G$ has no compact open subgroups. We must show that $G$ has a discrete infinite cyclic subgroup. As $C$ is noncompact by (1) and (2), we can assume $G=C$, that is $G$ is connected. Then (3) implies that $G$ contains a compact normal subgroup $K$ such that $G/K$ is a connected noncompact Lie group, and we can assume $K=1$, that is $G$ is a connected noncompact Lie group.
It suffices to prove that $Ad(G)$ has an infinite cyclic subgroup, that is, we can assume $G\subset GL_n(\mathbb R)$.
Say that an endomorphism $x$ of a real finite dimension vector space $V$ satisfies condition (C) if it is semisimple with imaginary eigenvalues, or, equivalently, if the subgroup $exp(\mathbb Z x)$ of $GL(V)$ is not infinite discrete.
Let $\mathfrak g$ be the Lie algebra of $G$, and $x$ be in $\mathfrak g$. If $x$ is nonzero and satisfies (C), the same holds for $ad(x)$, implying $Killing(x,x) < 0$. If it were so for all nonzero $x$ in $\mathfrak g$, then $\mathfrak g$, and thus $G$, would be compact.
[1] Willis, G. The structure of totally disconnected, locally compact groups. Math. Ann. 300 (1994), no. 2, 341-363.
http://gdz.sub.uni-goettingen.de/en/dms/load/img/?IDDOC=167209
[2] Willis, G. Totally disconnected, nilpotent, locally compact groups. Bull. Austral. Math. Soc. 55 (1997), no. 1, 143-146.