Say I have a countably infinite initial set $A$ with $A \subset [0,1]$. My set satisfies the property that $\inf_{x\in A} |x-y| = 0,\,\forall{y\in [0,1]}$ (For example $A$ can be the set of rational numbers). I have a "construction" that works in the following manner:
Step 1: Remove an arbitrary $x_1\in A$ from $A$ to come up with the set $A_1 = A - \{x_1\}$,
and for $n=2,\ldots,N$,
Step n: Remove an arbitrary $x_n \in A_{n-1}$ from $A_{n-1}$ to come up with the set $A_n = A- \{x_1, \ldots, x_n\}$.
It can be shown that for any $n$, we have $\inf_{x\in A_n}|x-y| = 0,\,\forall{y\in [0,1]}$ as well. That is, removing finitely many terms from $A$, we still get $0$. I cannot guarantee though that repeating the above process infinitely many times I still get $0$, as I may have removed all the elements of $A$. My first question is why this is happening? Or to better put it, how can I avoid making mistakes when going from a finite number of steps to a countably infinite number of steps?
Now, consider another "construction"; at step $n$ I just remove an arbitrary element $x_n$ from $A$ to come up with the set $A_n = A - \{x_n\}$. Similar to the previous case we have $\inf_{x\in A_n} |x-y| = 0,\,\forall{y\in [0,1]}$ for any $n$. But now, I guess, even if I repeat the steps of this new construction infinitely many times, I would still get $0$ as my final answer. How can I prove this? In general, I have a construction in which I repeat some well-defined steps infinitely many times; how can I verify the correctness of my final answer? (You may suggest that I need to find a set-limit $\lim A_n$ for this case, but it appears to me that I need not, as I am "sure" that the final answer should be $0$ in any case).