The unit interval [0,1] is compact. This follows from the Heine-Borel Theorem. Proving that theorem is about as hard as proving directly that [0,1] is compact. The half-open interval (0,1] is not compact: the open cover $(1/n,1]$ for $n=1,2,\mathrm{\dots }$ does not have a finite subcover.

Again from the Heine-Borel Theorem, we see that the closed unit ball of any finite-dimensional normed vector space is compact. This is not true for infinite dimensions; in fact, a normed vector space is finite-dimensional if and only if its closed unit ball is compact.

Any finite topological space is compact.

Consider the set $2^{\mathbb{N}}$ of all infinite sequences with entries in $\{0,1\}$. We can turn it into a metric space by defining $d((x_n),(y_n))=1/k$, where $k$ is the smallest index such that $x_k\ne y_k$ (if there is no such index, then the two sequences are the same, and we define their distance to be zero). Then $2^{\mathbb{N}}$ is a compact space, a consequence of Tychonoff’s theorem. In fact, $2^{\mathbb{N}}$ is homeomorphic to the Cantor set (which is compact by Heine-Borel). This construction can be performed for any finite set, not just {0,1}.

Consider the set $K$ of all functions $f:ℝ\to [0,1]$ and defined a topology on $K$ so that a sequence $(f_n)$ in $K$ converges towards $f\in K$ if and only if $(f_n(x))$ converges towards $f(x)$ for all $x\in ℝ$. (There is only one such topology; it is called the topology of pointwise convergence). Then $K$ is a compact topological space, again a consequence of Tychonoff’s theorem.

Take any set $X$, and define the cofinite topology on $X$ by declaring a subset of $X$ to be open if and only if it is empty or its complement is finite. Then $X$ is a compact topological space.

If $H$ is a Hilbert space and $A:H\to H$ is a continuous linear operator, then the spectrum of $A$ is a compact subset of $ℂ$. If $H$ is infinite-dimensional, then any compact subset of $ℂ$ arises in this manner from some continuous linear operator $A$ on $H$.

If $𝒜$ is a complex C*-algebra which is commutative and contains a one, then the set $X$ of all non-zero algebra homomorphisms $\varphi :𝒜\to ℂ$ carries a natural topology (the weak-* topology) which turns it into a compact Hausdorff space. $𝒜$ is isomorphic to the C*-algebra of continuous complex-valued functions on $X$ with the supremum norm.

Any locally compact Hausdorff space can be turned into a compact space by adding a single point to it (Alexandroff one-point compactification (http://planetmath.org/AlexandrovOnePointCompactification)). The one-point compactification of $ℝ$ is homeomorphic to the circle $S^1$; the one-point compactification of ${ℝ}^2$ is homeomorphic to the sphere $S^2$. Using the one-point compactification, one can also easily construct compact spaces which are not Hausdorff, by starting with a non-Hausdorff space.