[I realize I had misread the question, as I understood that the real subgroup required to be compact is the normal one. Since asking the question with the normal subgroup ($X$ in Dave's post) required to be compact seems much less trivial than the original question (answered by Dave), I'll include the full answer to the modified question.]
To avoid ambiguity, let me denote by $R$ the underlying additive group structure on the reals, without topology. So it's just a $\mathbf{Q}$-vector space of continuum dimension, so a group is isomorphic to $R$ iff it's abelian, torsion-free, and divisible, of continuum cardinal.
As mentioned in the comments, there exist compact group topologies on $R$.
However, I claim that if $T$ is a compact topology on $R$, then the group of automorphisms of the topological group $(R,T)$ is not transitive on $R-\{0\}$. This implies that the underlying group of $\mathbf{R}\rtimes\mathbf{R}$ cannot be endowed with a group topology making the normal $\mathbf{R}$ compact.
To prove the claim, write $K=(R,T)$. This is a compact abelian group, with no continuous homomorphism onto any nontrivial finite group. Hence its Pontryagin dual is a torsion-free abelian group $A$. If $I$ is a maximal free family in $A$, then the embedding $\mathbf{Z}^{(I)}\subset A$ induces a (continuous) surjection $K\to (\mathbf{R}/\mathbf{Z})^I$. Since $K$ has continuum cardinal, this implies that $I$ has cardinal less than continuum. Hence, noting that $A$ has the same cardinal as $I$, the cardinal of $A$ is less than continuum.
On the other hand, for each $x\in\mathbf{R}/\mathbf{Z}$, $x$ belongs to the image of some homomorphism $A\to \mathbf{R}/\mathbf{Z}$: this comes from basic Pontryagin duality: just find an infinite cyclic subgroup $\langle m\rangle$ in $A$, consider the homomorphism $m\mapsto x$, and extend it to $A$. As a consequence (using that $A$ has cardinal less than continuum), the number of possible images of homomorphisms $A\to\mathbf{R}/\mathbf{Z}$ is continuum. Now the group $\mathrm{Aut}(K)$ acts on this set of homomorphisms, and this action does not change the image, so the action of $\mathrm{Aut}(K)$ on $\mathrm{Hom}(A,\mathbf{R}/\mathbf{Z})$ has continuum many orbits. By Pontryagin duality, the action of $\mathrm{Aut}(K)$ on $\mathrm{Hom}(A,\mathbf{R}/\mathbf{Z})\simeq K$ has continuum many orbits, in particular it is not transitive on $K\smallsetminus\{0\}$.