Compactness of the closed interval [0,1] In general topology, a topological space is said to be compact, if every one of its open cover has a finite subcover.
However, I cannot see the compactness of the close interval [0,1] from the above definition.
To be a little specific,let us consider the following open cover for [0,1]: $C= \{[0,1/2),(1/3,3/4), (2/3,1]\}$. Now, the open interval (1/3,3/4) itself has at least one open cover (let's call it P) which does not have a finite subcover. We use P to cover the open interval (1/3,3/4). This gives a new open cover C' for the interval [0,1]. It looks like C' does not have an finite subcover, since C' includes P which does not have a finite subcover.

Of course I misunderstood something here. If somebody can catch my error it will be very helpful.
So you get an open cover by retaining $[0,1/2)$ and $(2/3,1]$
but replacing $(1/3,3/4)$ by a bunch $P$ of open sets where no finite
collection covers $(1/3,3/4)$.

You can do this.

But it is still the case that this new covering $C'$ has a finite
subcovering. Don't forget that $[0,1/2)$ and $(2/3,1]$
are still available. If we used both of these, all we have to do
is find a finite subset of $P$ that covers $[1/2,2/3]$. (We don't
need it to cover all of $(1/3,3/4)$.) As $[1/2,2/3]$ is compact,
then there will be such a finite subset.
This kind of thing (compactness of $[0,1]$) is typically proved via an appeal to the Heine-Borel theorem, which says that a subset of $\mathbb{R}^n$ is compact if and only if it is closed and bounded.  The subset $[0,1]$ is closed and bounded and therefore compact.