show/hide this revision's text 2 Corrected small error.

Here's a proof.

Lemma: Suppose $G$ is a (word-)hyperbolic group acting properly discontinously, cocompactly and faithfully on a C⁢A⁢T⁢(0) space $X$. Then there is a uniform bound $R_0$ on the width $R$ of isometrically embedded flat strips $\mathbb{R}\times [0,R]$ in $X$.

Proof: If not then, by cocompactness, there exist nested strips flat discs of width diameter tending to infinity. Their union is an embedded copy of $\mathbb{R}^2$, which contradicts hyperbolicity. QED

Now suppose that walls $W,W'$ have stabilizers with infinite intersection. Then that intersection is an infinite word-hyperbolic group (since it is quasiconvex in $G$) and so contains an element $\gamma$ of infinite order. Because $W,W'$ are convex and so themselves CAT(0), each contains an axis $l_W,l_{W'}$ for $\gamma$. By standard facts about CAT(0) spaces (see Bridson--Haefliger), any two axes bound a flat strip. Therefore, by the lemma, $l_W$ and $l_{W'}$ are at distance at most $R_0$, as claimed.
show/hide this revision's text 1

Here's a proof.

Lemma: Suppose $G$ is a (word-)hyperbolic group acting properly discontinously, cocompactly and faithfully on a C⁢A⁢T⁢(0) space $X$. Then there is a uniform bound $R_0$ on the width $R$ of isometrically embedded flat strips $\mathbb{R}\times [0,R]$ in $X$.

Proof: If not then, by cocompactness, there exist nested strips of width tending to infinity. Their union is an embedded copy of $\mathbb{R}^2$, which contradicts hyperbolicity. QED

Now suppose that walls $W,W'$ have stabilizers with infinite intersection. Then that intersection is an infinite word-hyperbolic group (since it is quasiconvex in $G$) and so contains an element $\gamma$ of infinite order. Because $W,W'$ are convex and so themselves CAT(0), each contains an axis $l_W,l_{W'}$ for $\gamma$. By standard facts about CAT(0) spaces (see Bridson--Haefliger), any two axes bound a flat strip. Therefore, by the lemma, $l_W$ and $l_{W'}$ are at distance at most $R_0$, as claimed.