> There is no proof that will ever satisfy a person dead-set against this.

Yes there is. There is a proof that uses only fundamentals of first year university analysis. When you see

0.99999....

this can be written as an infinite sum

\sum_{i=0}^\infty 0.9 x 10^{-i}.

Truncate the sum and set

S_n = \sum_{i=0}^n 0.9 x 10^{-i}

and now simply use the rules of arithmetic progressions to get the limit out:

0.1 S_n = \sum_{i=0}^n 0.9 x 10^{-i-1}

S_n - 0.1 S_n = 0.9 - 0.9 x 10^{-n-1}

0.9 S_n = 0.9 ( 1 - 0.1^{n+1} )

S_n = 1 - 0.1^{n+1}

Now let n tend to infinity to find the limit, which is 1.

You don't need to imagine infinity to do any of this.