Does a cover have to be infinite
Just cover it with one-element sets: $\{\{x\}, x\in[1,2]\}$. This is uncountable infinite cover and you cannot omit even a single one of them!
A cover does not have to be infinite, and the example you gave is in fact a finite cover. A cover of a set $S$ is simply a collection $\mathcal{F}$ of sets such that for any $x \in S$, there is at least one $A \in \mathcal{F}$ with $x \in A$. There can be any number of sets in $\mathcal{F}$, and they don't even need to be disjoint. But there are a few issues with the example you gave, and how your finite cover relates to it:
- Since the cover you gave is finite, it does have a finite subcover: the cover itself.
- What you want to do is find a particular infinite cover $\mathcal{F}$ such that there does not exist a finite subset of $\mathcal{F}$ that is also a cover.
- In most cases (i.e. for showing compactness) you want $\mathcal{F}$ to be an open cover, that is, you want the sets in $\mathcal{F}$ to all be open.
It's actually very simple. Just take $[1,1.1],[1.9,2]$ to be in the cover. Now what is left over is $(1.1,1.9)$, which is not closed, so not compact. Hence, there is an open cover not having a finite subcover. For example, $(1.1 + \frac 1n, 1.9-\frac 1n)_{n \in \mathbb N}$ would do. Note that these are nested, so the union of the sets of any finite subcover, is just the largest set of that cover, but none of the sets are equal to $(1.1,1.9)$, so this qualifies.
So, $\mathcal U = \{[1,1.1],[1.9,2],\left(1.1 + \frac 1n,1.9-\frac 1n\right)_{n \in \mathbb N}\}$ works as a infinite cover having no finite subcover. Note it does contain non-open sets, as you remarked earlier.
EDIT : This construction is of a "countable" cover not having a finite subcover. For a nice trivial example, the other answer does well.