Convex hull of a connected subset on a complete surface of non-positive curvature Let $S$ be a simply connected surface, possibly with boundary components, with a smooth complete metric of non-positive curvature. Let $X\subset S$ be a closed connected subset. I would like to know if the following procedure produces a convex subset of $S$ that contains $X$:
Construction. For any two points $x_1,x_2\in X$ let $[x_1,x_2]$ be the length minimizing path connecting $x_1$ and $x_2$. Now define $\Omega(X)$ as the union in $S$ of all such paths $[x_1,x_2]$ over all  $x_1,x_2\in X$.
Question. Is $\Omega(X)$ convex in $S$?
PS. I have to add some information. A set $C$ is convex if for any two $x_1,x_2\in C$ and any minimizing path $[x_1x_2]$ $C$ contains it. 
 A: Yes, it is true. 
First note that $S$ is CAT[0] space. This can be proved along the same lines as in "The intrinsic geometry of a Jordan domain" by Richard Bishop (you may also check the proof of Theorem 4.4.1 in our paper).
Further note that any geodesic in $S$ can be extended to a maximal geodesic (infinite or with the ends on the boundary).
This is a standard lemma — geodesics in a CAT[0] space might terminate only at the points which admit contactable punctured neighbourhood.
Assume $\Omega(X)$ is not convex, choose a point $w\notin \Omega(X)$ which lies between $u,v\in \Omega(X)$.
Assume $u\in[xy]$ and $v\in [zt]$ for $x,y,z,t\in X$
Let $\gamma$ be a maximal extension of the geodesic $[uv]$.
Since $X$ is connected, it has to intersect $\gamma$.
Since $w\notin \Omega(X)$, the set $X$ might intersect $\gamma$ only on one side from $w$.
WLOG we can assume that it happens on the side of $u$.
Consider the maximal extension $\sigma$ of $[zw]$.
Let $s$ be a point in $\sigma\cap X$ which lies in the closure of connected component of $t$ in the complement $X\backslash \sigma$.
It remains to note that $w\in [zs]$
