Here is the proof in the case of locally symmetric spaces; you will have to combine it with "rank rigidity theorem" and Igor's comments to get a complete proof.

Let $G$ be a semisimple Lie group, $K\subset G$ a maximal compact subgroup, $B\subset G$ a Borel subgroup; $X=G/K$ is the symmetric space of $G$. Let $r$ denote the real rank of $G$, i.e., the rank of $X$. Let $\Gamma\subset G$ be a lattice. Consider the action of $\Gamma$ on $G/B$, the complete flag manifold of $G$, also known as Furstenberg boundary of $X$. (Geometrically, one thinks of elements of $G/B$ as Weyl chambers at infinity in the Tits boundary of $X$.) The action of $\Gamma$ on $G/B$ is known to be minimal, i.e., every orbit is dense. You should be able to find a proof in Mostow's book on strong rigidity. The other ingredient you need is a theorem of Prasad and Raghunathan (Annals of Math., 1972) that $\Gamma$ contains a semisimple subgroup $\Lambda$ isomorphic to ${\mathbb Z}^r$: Such subgroup stabilizes (unique) maximal flat $F$ in $X$ and acts on $F$ cocompactly. It is elementary that end-points of axes of elements of $\Lambda$ are dense in the sphere $S^{r-1}$, the geometric boundary of $F$ (rational directions are dense in the unit sphere).

Now, we can put all this together: Given a point $\xi$ in the geometric boundary of $X$, let $\sigma$ be a Weyl chamber at infinity containing $\xi$. Let $\tau$ be a chamber at infinity stabilized by the abelian subgroup $\Lambda\subset \Gamma$ as above. Let $\gamma_n\in \Gamma$ be a sequence such that $\lim_n \gamma_n(\tau)=\sigma$. Since ends of axes of elements $\lambda$ of $\Lambda$ are dense in $\tau$, their images under the sequence $\gamma_n$ will accumulate to $\xi$ as well. These images are ideal points of axes of conjugates $\gamma_n \lambda \gamma_n^{-1}$. qed