Is a ball noncompact? A compact manifold usually refers to "a manifold without a boundary", for example the usual 2-sphere $S^2$. 
What about a manifold with a boundary? Intuitively, I think such an example, e.g. a ball in $\mathbb{R}^2$ with $S^2$ as a boundary is "compact", but it does have a boundary. So, why does compact usually connote something without a boundary?
The basic definition of compactness of a topological space X is that every open cover of X has a finite subcover. Its unclear that the boundary of the ball by itself is a compact manifold, but when supplemented with the volume inside, becomes only the boundary of the manifold.
 A: The general topological notion is usually not needed, and "closed and bounded" works instead.  Bounded means you don't have to worry about a bunch of balls of size one failing to cover it no matter how many (finitely many) you have. Being closed means you don't have to worry about there being some boundary that wasn't included.
If you didn't include the boundary then for every point you could restrict your neighborhood to only go up to half way towards the boundary, so the stuff closer to the boundary is obviously excluded and it is like an infinite regress, every neighborhood has a point closer to the boundary that isn't in the neighborhood, so a finite number of neighborhoods doesn't suffice.
This doesn't prove that closed and bounded means compact, but unbounded is not good for compactness and not closed is not good for compactness.
Having a boundary is however a separate problem for physics even if you include the boundary. Because in general relativity you want every event to have a neighborhood where everything looks like Special Relativity and there is no boundary (included or excluded) in special relativity.
As for the spherical surface of a ball, that is obviously compact. It is a subset of $\mathbb R^3,$ it is bounded, and it is closed. And while the closed unit ball is compact, it has a boundary, $S^2,$ so violates a basic principle that different points should look the same. The points on the edge look different than the points on the inside, even when you look locally in space and time and even beyond the relational aspects of fields here versus fields nearby, it is fundamentally is different. And that just breaks other more basic principles in Physics.
So it is often already excluded so was already not being considered. And is unrelated to compactness.
That said, you might consider something with boundary. For instance a time forward solution to an initial value problem might have the initial values as a boundary. Or a boundary value problem. In a sense if that surface is already special because you are specifying rather than solving then it isn't as big a deal. Because you can imagine that you had specified elsewhere and gotten that as well as the rest.
A: A manifold is a space $X$ such that each point $p\in X$ has  neighborhoods which are homeomorphic to open euclidean balls. In particular the neighborhoods of all points $p\in X$ "look alike". If the manifold $X$ is  a compact  topological space then  $X$ is called a closed manifold (this wording is somewhat unfortunate, see below). The prime examples of closed manifolds are spheres $S^d$ and tori $T^d:={\mathbb R}^d/{\mathbb Z}^d$. A parabola in the plane is a non-compact onedimensional manifold.
Now there are interesting geometrical objects which are not manifolds, e.g. closed balls $$B^d:=\bigl\{x\in{\mathbb R}^d\>\bigm|\>|x|\leq 1\bigr\}$$ (note that here the word "closed" has  the usual topological meaning, referring to a subset of some larger space). The space $B^d\subset{\mathbb R}^d$ is not a manifold: The points $x\in S^{d-1}\subset B^d$ are "special", insofar as they don't have neighborhoods of the required structure. But $B^d$ can be considered as a manifold with boundary, whereby the exact definition involves a lot of technical conditions.
A: Note well that the answer is necessarily positive only in finite dimensions, I.e. Balls in R^n. 
In infinite dimensions, it depends on the topology used. For example, in infinite dimensional Hilbert space, the unit ball is non-compact in the topology induced by the norm. This can be seen as a basis can be viewed as a sequence without accumulation point. If it were compact, Bolzano weierstrass would imply the existence of an accumulation point. In weak topology, however, the ball is compact and the sequence converges to 0. 
