I hope you will bear with me and excuse me if my question is kind of obvious to many of you. In an $n$-manifold say we have $x,y$ in it. Could we always find an open $n$ dimensional ball contained in the manifold such as both points belong to it?
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You need at least that the manifold is path-connected. I think the following works: given a path $p : I \to M$ with no self-intersections from $x$ to $y$ we can cover every point in the image of $p$ by a small open subset homeomorphic to the open $n$-ball. By compactness this cover has a finite subcover, and by choosing the balls small enough (not in the sense of a metric but in the sense that they only intersect each other when consecutive) their union should also be homeomorphic to an $n$-ball.
Edit: Ryan Budney's explanation in the comments shows I am being naive. It is not obvious that the existence of a path implies the existence of a particularly nice path without additional work.
To complement Qiaochu's answer, here is why this is impossible when $M$ is not path-connected (note that manifolds are connected if and only if path-connected, this is Prop 1.8 in Lee Smooth Manifolds).
If $x,y\in M$ are in different connected components of $M$, then any open set of $M$ that contains $x$ and $y$ would necessarily be disconnected, which an open $n$-ball is not.
Another argument would be to put a complete Riemann metric on the manifold -- this can be done fairly easily, for example if the manifold was a closed subset of Euclidean space (can be done via the Whitney embedding theorem) then the induced metric is complete by Hopf-Rinow: http://en.wikipedia.org/wiki/Hopf-Rinow_theorem
Okay, so now the manifold is geodesically complete, so given any two points $p,q \in M$ take a shortest geodesic segment connecting $p$ to $q$. This is an embedded arc by design. Now a little regular neighbourhood of that arc is what you're looking for.