How to show that an infinite decimal is equal to a unique real number? 
I don't understand how the proof above shows that two distinct real numbers correspond to different infinite decimal.
All I got out of the explanation is given any two distinct real numbers $a$ and $a'$, we can always pick an integer $n$, such that $\frac{1}{10^n}<|a-a'|$, which to me means that there is always a number that is smaller than the absolute difference between the two given real numbers $a$ and $a'$. What is the significance of this and how does it relate to the objective of showing that two distinct real numbers correspond to different infinite decimal? I am also puzzled at the conclusion "...will certainly differ by the nth decimal place"
I tried to follow workings with some example numbers say $a=6$ and $a'=6.1$ but still did not make any progress. 
 A: It does not just mean that there is always a number that is smaller than the absolute difference between the two given real numbers.  Rather it means that there is such a number of the form $1/10^n$, so that digits in the $n$th decimal places in the two numbers differ from each other.
PS: For example, two numbers differing in the $5$th place and not before are two numbers that differ by $1/10^5$.  Two numbers differing by at least $1/10^5$ are numbers that differ at or before the $5$th place after the decimal point.
\begin{array}{cccccccccc}
& 3 & . & 1 & 4 & 1 & 5 & 9 \\
-& 3 & . & 1 & 4 & 1 & 5 & 8 \\
\hline
& 0 & . & 0 & 0 & 0 & 0 & 1 & = & 1/10^5
\end{array}
A: To go through your example:
We note that $|a - a'| > 1/10^{2}$.  So, according to the proof, the decimals should differ by the $2$nd decimal place.
Indeed, this is the case.  We note that the $0,1,$ and $2$ digits of the representation $a$ must be could be $6,0,0$ since
$$
6 \leq a \leq 6+1 = 7\\
6.0 \leq a \leq 6.0 + 0.1 = 6.1\\
6.00 \leq a \leq 6.00 + 0.01 = 6.01
$$
So, is the same true for $a'$?
$$
6 \leq a' \leq 6+1 = 7 \quad \text{. . . good so far}\\
6.0 \leq a' \leq 6.0 + 0.1 = 6.1 \quad \text{. . . nothing wrong yet}\\
6.00 \leq a' \leq 6.00 + 0.01 = 6.01 \quad \text{. . . this isn't true!}
$$
So, we see that the two numbers can't have the same decimal representation.
