Does a subcover need to be a proper subcover?

Question

Definition: A Space $\mathbb{X}$ is said to be Compact if every Open Cover has a Finite Subcover.

My Question:
Suppose, Space $\mathbb{X}$ has an Open Cover $A$ whose finite subcover is $A_{1}$.
Now, $A_{1}$ becomes a open cover of space $\mathbb{X}$. Suppose that $A_{1}$ has a finite subcover $A_{2}$. Again, Repeat the process.
After fintely many steps of this process, we will arrive at a stage when, there will be no finite subcover for the previously made cover. (Say, the process ends at Nth step. So, there is no finite subcover at (N+1)th step. )
So, By Definition, I conclude that: There is No space $\mathbb{X}$ which is compact.

Answer 1

This community wiki solution is intended to clear the question from the unanswered queue.

A cover $\mathfrak{C}$ of $X$ is a family $(C_\alpha)_{\alpha \in A}$ of subsets $C_\alpha \subset X$ such that $\bigcup_{\alpha \in A} C_\alpha = X$. A subcover $\mathfrak{C}'$ of $\mathfrak{C}$ is any subfamily $(C_\alpha)_{\alpha \in A'}$ with $A' \subset A$ such that $\bigcup_{\alpha \in A'} C_\alpha = X$. In particular, $A' = A$ is allowed. This is equivalent to $\mathfrak{C}'= \mathfrak{C}$.