Shortest paths in Alexandrov spaces

Let $X$ be an Alexandrov space with curvature bounded from below (if necessary, $X$ might be assumed to be finite dimensional or even compact).

Question 1. Is it true that every point of $X$ has a neighborhood $U$ such that any two points from $U$ can be connected by at most one shortest path (which does not have to be contained in $U$)?

Question 2. In the previous question, can one choose $U$ to be geodesically convex, i.e. for any two points from $U$ any shortest path between them (if it is not unique) is contained in $U$?

Remark. Of course, if $X$ is a smooth Riemannian manifold then the answers to both questions are 'yes'.
• I am afraid, the vertex $p$ of the cone $X$ over the circle of length less than $2\pi$ has no such neighborhood $U$. – valeri Apr 1 '15 at 9:50