Is $[0, 1]$, with the discrete topology, compact?

Question

Is $[0, 1]$, with the discrete topology, compact?

This was on my exam today and I have no idea how to approach it. I thought of using $\mathcal{A}=\{[x, x+0.1]\}$ as an open cover of $[0, 1]$ but it didn't work.

Answer 1

No infinite set $X$ endowed with the discrete topology is compact, because $\bigl\{\{x\}\,|\,x\in X\bigr\}$ is an open cover with no finite subcover.

Answer 2

Every infinite set $X$ with the discrete topology isn't compact. This is a consequence of the fact that $$ \bigcup_{x \in X}\{x\} $$ is an open cover of $X$ which has no finite subcover.

Answer 3

In the discrete topology every set is open. That includes the singleton sets. $[0, 1]$ is covered by the set of all singleton sets. Can it be covered by a finite set of singleton sets?