algebra precalculus - Why does $0,bar{9}$ equal $1$?

I am finding hard to understand why $0,99999..... = 1$

I have the following proof:

Let $x$ be $0,9999...$

then $10x = 9,999...$

So $10x - x = 9,999 - 0,9999$

$9x = 9 \rightarrow x = 1$

From a philosophical respective, it does seem legit to me that if the decimal form of the number is never ending, then, at infinity and beyond it "tries" to reach 1, so it's limit to infinity equals 1.

My objection is that:

Consider the set $S = \left\{ 0, 1, 2, 3, ..., n \right \}$

What is the possiblity that from the set S we obtain the number $2$?

$P = \frac{P(a)}{P(S)}$ = $ \frac{1}{n}$

so from there we can see that the possiblity is very low, $0.000...1$ , which again we can consider to be $0$ since :

$ \frac{1}{n}$ = $ \frac{1}{\infty}$ = $0$

But again, if we accept that $P = 0$, then there is no possibility that we can select the number 2 from the set $S$, which is false.