 Solution

Let $[u,v]$ be one side of the geodesic quadrilateral, and draw a diagonal from $u$ to the opposite vertex. A point $p\in [u,v]$ lies, by $\delta$-thinness of the first geodesic triangle, within $\delta$ either of an adjacent side or of the diagonal. In the latter case, $\delta$-thinness of the second triangle places the nearby point of the diagonal within another $\delta$ of one of the other two sides. The triangle inequality then places $p$ within $2\delta$ of the remaining three sides. The argument applies to every side, proving the geodesic quadrilateral in a hyperbolic metric space bound.