Can anyone help me show directly that [a,b] is covering compact? By directly, I meant that I am not allowed to show that [a,b] is sequentially compact which implies covering compact, by a theorem we learned.

Hint:

If we let U be an open covering of [a,b] and let C={ x in [a,b]: finitely many members of U cover [a,x]}. Try to show that b is in C by the least upper bound property.

I really have trouble with the covering compact concept. We have the definition of covering compact as : M is a metric space, A in M is covering compact if every covering U of A has a finite subcovering.