What is wrong with this argument that closed interval [0, 1] is not compact?

By the definition it's compact. You misunderstood and the proof is incorrect.

You should try to prove that if you take an arbitrary infinite cover of E, you can find within that cover a finite sub-cover. What you did was: you picked one particular infinite cover of E and then you showed that it has an infinite sub-cover. That doesn't do the work here.

Your "proof" is wrong.

The definition says that for every cover there exists a finite subcover. In your post there is an example of an infinite subcover of $\{U_\alpha\}$.

By following your reasoning every subset $U$ of a topological space would be non-compact: pick a "redundant" infinite open cover of $U$ (you can always find it) and remove one set. You have an infinite subcover.

As said in the other answers, you misunderstood a point of the definition.

It is actually tricky to prove that $[0,1]$ is compact. You have to consider an arbitrary cover of $[0,1]$, let's call it $\mathcal{U}$. Then, consider the set $A=\{x\in [0,1]$ such that $\mathcal{U}$ has a finite subcover covering $[0,x]\}$. The proof itself is done in three steps: First, prove that $A$ is non empty. Then prove that sup$A\in A$. And finally, prove that sup$A=1$. Maybe it doesn't look so hard, but the third step is not trivial.