What is wrong with this argument that closed interval [0, 1] is not compact? I am a student majoring engineering.
I am studying real analysis with textbook 'Measure and Integral' by Wheeden and Zygmund.
This book defined compact like the following:

$E$ is compact if every open cover of $E$ has a finite subcover.

By the definition $[0, 1]$ is not compact.
However, by the Heine-Borel theorem, $[0, 1]$ is compact.

Let me prove why $[0, 1]$ is not compact by the definition.
According to the definition, it is enough to show an open cover of $E$ having infinite subcover.
If $C=\{U_\alpha:\alpha\in \mathbb{N}\}$ is an indexed family of sets $\displaystyle U_\alpha=\left(-1+\frac{1}{n}, 2\right)$,
then $C$ is a cover of $[0, 1]$ because $\displaystyle [0,1] \subseteq \bigcup\limits_{\alpha  \in \mathbb{N}} {\mathop U\nolimits_\alpha  }$.
This $C$ has infinite subcovers like $C=\{U_{2\alpha}:\alpha\in \mathbb{N}\}$
Can someone teach me what is my fault?

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.
