This is the definition I use for compactness: Let $X$ be a topological space. We say that $K \subset X$ is a compact set if:

1. $K$ is not empty
2. For any arbitrary open sets $U\_i \subset X\_{i∈I}$ whose union contains $K$, one can find a finite number of these open sets such that their union contains $K$

I want to demonstrate that $K\_1 = [1,2]$ is compact and $K\_2 = [1,2)$ is not, being $X=\mathbb{R}$ equipped with the usual topology.

For both cases I start saying that the space of all open covers $S\_i$ exists for each $K\_i$. Then I build a set $U\_i = (0,2 - 1/i), i > 0$ and build a collection of sets $A\_n = \{ U\_i \mid 1 \le i \le n \}$

* For $K\_2$, I recognize that:
	1. For any $x \in K\_2$, I can always find a collection (of sets) $A\_N$ such that the union of its members $U\_i$ contains such $x$, so basically $K\_2 \subset\bigcup U\_i \in S\_2$ .
	2. If I try to join a finite amount of $U\_i$, say up to $i = M$, I can find a $x \in K\_2$ such that $x \notin \bigcup\_{i=1}^{M} U\_i$.
	3. Then, I have found a collection of open sets whose union contains $K\_2$ (1) but it is impossible to find a finite collection of those sets whose union still contains $K\_2$ (2).
* For $K\_1$, I recognize that I can find $x$ such that I would never find it on any $A\_n$, this is $x=2$. If I build the collection of open sets $A'\_i = \{U\_i ,B\_\epsilon(2) \mid i > 1\}$, where $B\_\epsilon(1), \epsilon > 0$ is a (open) ball around $2$ of radius $\epsilon$.
	1. First I show that it is possible to cover $K\_1$. For any $x \in K\_1$, if $x = 2$, then $x\in B\_\epsilon (2) \subset A'\_i$ for any $A\_i$. If $1 \le x < 2$ one can always find a collection (of sets) $A\_N$ such that their union of its member $U\_i$ contains such $x$, so, for any $x \in K\_1$, it's possible to find a collection (of sets) $A'\_N$ whose union of its members contains $K\_1$: 
	\begin{eqnarray} K\_1 \subset \left(\bigcup U\_i \right) \cup B\_\epsilon \in S\_1, \text{for any $\epsilon$} \end{eqnarray}
	2. Now we prove that it is not necessary to have all $U\_i$. Given $M$, and $x \in K\_1$, one can find a $x \notin \bigcup\_{i=1} ^ M U\_i$ and for any $x \notin \bigcup\_{i=1}^M U\_i$ there is a $B\_\epsilon (2)$ such that $x \in B\_\epsilon (2)$ for some $\epsilon \ge d$. Basically, I can find a finite collection of open set $A'\_N =\{A\_N, B\_d(2)\}$ whose union of its memebers contains $x$
	3. We've shown that for any arbitrary collection of open sets (is this really arbitrary?) such that their union contains $K\_1$, it is possible to find a finite collection of those open sets such that their union still contains $K$.

I want to know if there are mistakes, specially in the case for $K\_1$.