The problem is: "Prove that $[0,1]$ is compact using the definition of compactness"
So we cannot use the Heine Borel Theorem which states that any closed bounded set of $\mathbb{R}^n$ is compact. We have to use the definition of compactness which is that for any open cover of the set, there exists a finite subcover.
Consider an open cover $G$ of $[0,1]$. Then $0$ and $1$ are interior points of an open set. Hence there exist neighborhoods $N_0$ and $N_1$ around $0$ and $1$ respectively (with radius $\varepsilon$) such that $N_0 \subset G$ and $N_1 \subset G$. Then consider the set $ E = (\varepsilon/2, 1 - \varepsilon/2)$. The union $N_0 \cup E \cup N_1$ is then a open cover of $[0,1]$ which is also a subset of $G$, hence it is a finite sub cover of $G$. Therefore $[0,1]$ is compact.
But I have looked online of proofs and I get completely different arguments. Am I wrong here? I feel like I am. Can anyone point to the incorrectness?
Hint: if $G$ is an open cover for $[0,1]$, and let $A=\{a \in [0,1]: [0,a] \text{ has a finite subcover from } G\}$. Trivially $0 \in A$, as $0$ is covered by some element of $G$. So $a_0 = \sup A$ exists. (lub property of $\Bbb R$). Try to reason why $a_0 < 1$ cannot happen.