Earlier quoted context omitted.
Regarding 4, don't believe anyone who claims it's an intuitive result; it is simply a properly of real numbers that you cannot have nonzero infinitesimals and that any two distinct numbers have a number between them that is not equal to either (infinitely many in fact). You can construct alternate number lines that do allow nonzero infinitesimals and then .9999... actually is not equal to 1 under that number line; th…
There do exist number systems where .999.../=1, however, they are not a strict superset of the reals. If it were, then any operation involving only real numbers would behave identicly to the real number system. Also, this property is not a mere convention, but rather a nessasary result of what we want the number line to be. For example, assume that X<Y. Consider Z=(X+Y)/2. Z=X/2+Y/2. X<Z<Y. I have just shown, using b…
To clarify my convention comment; it is only convention that mathematicians have decided that any given set of properties are useful or interesting to be used pervasively and alternate number systems are not. To any person who is asking why .9 repeating is 1 those reasons are entirely outside the scope of their knowledge and so entirely unrelated to the question; is it not true that the answer is "based on the axioms of which this number line is created due to complex reasons that you can't possibly know at this stage, this is effectively decided to be true as an axiom".
If one of these alternate number lines was in common use and reals were only uncommonly used then I'm pretty sure there would be tons of people saying "why is .99.. NOT equal to 1 in ($reals-replacement)" and all the people who have been taught it would sigh condescendingly because they believe it is just an intuitively obvious feature of numbers and not simply an axiom of the number line they are choosing to use. I'm interested in whether you disagree with that sentiment.
Re: your proof, trivially could be refused based on the other number line not being closed over division, then there would simply be no such number Z=(X+Y)/2 for X=.999.. and and Y=1. Then the proof where be like those silly ones where someone uses a /0 and arrives at a contradiction. Actually it seems that reals themselves already are not closed over division since x/0 doesn't result in a real.