A compact open set Is there an open set which is compact ? I would say that $\emptyset$ is an open set compact because it's bounded and closed too. Is it correct ?
 A: The empty set is certainly compact, all finite spaces are.
A compact subset of a Hausdorff space (for example a metric space) is always closed, so your set would be closed and open. Now that can happen, but in a connected space the only open and closed subsets are the empty set and the space itself.
A: If you look at $X=[0,1]\cup(4,5)$ as a topological subspace of $\Bbb R$, then $[0,1]$ is still compact, and it is open since it is equal to $(-1,2)\cap X$.
Going to a completely different direction, if you look at a set with the co-finite topology, then every set is compact. In particular every co-finite set, which is open. If the set is finite, this is just the discrete topology, but if the set is infinite, then this ends up being a stranger topology than you might be used to.
A: To give you an example which isn't pathological, and is very useful 'in the wild', the study of $p$-adic numbers gives rise to an example of compact open sets. 
I won't explain their construction in explicit detail here, but I will sketch it loosely. 
 As motivation, $p$-adic numbers belong to nice algebraic objects (rings and fields), but we're also able to do topology and even 'calculus' on them, so they play an important roll in number theory.  
For any prime $p$, one may form a set (in fact, a field) $\mathbb{Q}_p$ called the $p$-adic numbers.  This is obtained in a similar way that we get the real number from the rationals $\mathbb{Q}$, by 'completing' the set with respect to some notion of distance. (This notion of distance has to do with divisibility of the prime $p$).  
Inside $\mathbb{Q}_p$ lies a set (in fact, a ring) $\mathbb{Z}_p$ called the $p$-adic integers.  It turns out that $\mathbb{Z}_p$ is both an open set and compact, a feature which is as interesting as it is useful.    
A: Consider a set $X$ with more than two elements, equipped with discrete metric $d(x,y)=1$ if $x \not = y$ and $d(x,y)=0$ if $x=y$. I will leave it to you as an exercises to prove that a set is compact if and only if it is finite. Pick an element $a \in X$. Consider the ball $D(a,\frac{1}{2})$. Clearly,$D(a,\frac{1}{2})=\lbrace a \rbrace $ and hence $\lbrace a \rbrace $ is open and compact at the same time.
A: In a proper space a closed bounded set is compact but in general topological space it is not true that a closed bounded set is compact. There are plenty of example like spaces of infinite sequence. Take two disjoint open bounded close disc give it a subspace topology of $R^2$ you will have open set which is compact.
