Are there compact manifolds without boundary? Based on this question I'd like to know: Are there compact (sub)manifolds without boundary in $\mathbb{R}^n$ ? Because, as that question shows, the topology of the manifolds has to be the trace topology; thus compact subspace (in particular, compact manifolds) are characterized by the Heine-Borel theorem: They are precisely those sets in $\mathbb{R}^n$ that are closed and bounded.
But, as far as I know (and I'm just starting to read about manifolds and haven't got a good grasp on the formal definitions yet), manifolds without boundary aren't closed, so they can't be compact ?
 A: $\newcommand{\Reals}{\mathbf{R}}$Comparing this question with your linked question, the central issue seems to be the term "closedness", which perhaps feels bothersome because manifolds are unions of open sets.
If that's really the question, the resolution comes down to "relative topology", how "open" and "closed" are defined for subsets of $\Reals^{n}$.
If $X \subset \Reals^{n}$ is an arbitrary non-empty set, we say $A \subset X$ is (relatively) open in $X$ if there exists an open set $U \subset \Reals^{n}$ such that $A = X \cap U$. Relatively closed sets are defined similarly.
In particular, every set $X$ is both (relatively) open and closed as a subset of itself: The set $U = \Reals^{n}$ is both open and closed, and $X = X \cap \Reals^{n}$.
Now let's take an example of a compact manifold in $\Reals^{n}$, such as the $(n - 1)$-sphere
$$
S^{n-1} = \{x \text{ in } \Reals^{n} : \|x \| = 1\}.
$$
As the zero set of a continuous function $f:\Reals^{n} \to \Reals$, the sphere $S^{n-1}$ is closed in $\Reals^{n}$. Certainly, $S^{n-1}$ is not open in $\Reals^{n}$; in fact, $S^{n-1}$ has empty interior in $\Reals^{n}$.
As noted above, however, $S^{n-1}$ is both open and closed as a subset of itself. There's no contradiction because "open" and "closed" are relative concepts. (Remarkably and not completely obviously, "compactness" of $X$ is an intrinsic concept, not depending on how $X$ is viewed as a subset of a larger universe.)
Going back to the sphere, the hemispheres
$$
H_{i}^{+} = \{x \text{ in } S^{n-1} : x_{i} > 0\},\qquad
H_{i}^{-} = \{x \text{ in } S^{n-1} : x_{i} < 0\}
$$
are (relatively!) open subsets of $S^{n-1}$ (why?) constituting a covering by coordinate charts. It follows that $S^{n-1}$ is a compact manifold in $\Reals^{n}$.
(Incidentally, this is an "inefficient" covering of the sphere; stereographic projection allows the sphere to be covered with two coordinate charts, each covering the complement of one point.)
A: Well there are several examples. @Dorebell mentioned one example.
Here are some other compact manifolds without boundary:


*

*Torus

*Double Torus

*Klein bottle

A: The Riemann sphere is a compact complex manifold without boundary.
A: Yes; there are some. If $M$ is a compact, oriented $n$-manifold without boundary, then there must be some $n$-forms which do not arise from taking the exterior derivatives of $n-1$-forms. 
If $M$ is pseudo-Riemannian, so we have a Hodge star to work with, this tells us that we always have some functions on $M$ which are not the divergence of any vector field on $M$.
A: A compact manifold without a boundary is called a closed manifold so it is certainly an important class of manifolds.
Example: A $n-1$sphere ${\cal S}^{n-1}$ is a closed manifold, but the unit ball enclosed $\{\mathbf{x}\in \mathbb{R}^{n}\|\mathbf{x}\leq 1\|\}$ is a campact manifold with its boundary being ${\cal S}^{n-1}$. 
