Proving: If $|x-y|<\frac{1}{n}$ for every natural $n$ then $x=y$ It's really basic but I am trying to prove:
$$
\forall x,y\in \mathbb{R}.(\forall n\in \mathbb{N}. |x-y| <\frac{1}{n} \Rightarrow 
x=y)$$
I tried to prove it by proving that that if $$A= \left\{ \frac{1}{n} \ \middle| \ n\in\mathbb{N} \right\}$$
then the $$\operatorname{inf}(A) =0$$ but I couldn't find a way to use this after I proved it.
Unfortunately, I am not allowed to use limits here. 
 A: Suppose $x\neq y$, then
$$\vert{x-y}\vert>0$$
by the Archimedian property of $\mathbb{R}$ there exists $n\in\mathbb{N}$ such that:
$$n\vert{x-y}\vert>z,\quad \forall z\in\mathbb{R}\ \text{with}\ z>0$$
take in particular $z=1$, 
so there exists $m\in\mathbb{N}$ such that:
$$m\vert{x-y}\vert>1$$
which gives
$$\vert{x-y}\vert>\frac{1}{m}$$
this is a contradiction.
A: If $x\ne y$ then $|x-y|=d>0$ 
As $\frac{1}{n}\to 0$ when $n\to\infty$, then for any given $d>0$ there exists an index $n\in\mathbb{N}$ such that $|x-y|<\dfrac{1}{n}<d$. 
This is a contradiction. Therefore must be $x=y$.
Hope this helps
A: Let $x$ and $y$ be any two distinct real numbers with $y \lt x$. Then, there exist rational numbers $p$ and $q$ satisfying $y \lt p \lt q \lt x$. If $n$ is a common denominator for $p$ and $q$, then $|q - p| \ge 1/n$. But then of course  $|x - y| \ge 1/n$. 
A: We have:
$$\forall n \in \mathbb{N}: 0 \leq |x-y| < \frac{1}{n}$$
The squeeze theorem yields that $|x-y| = 0$, or equivalently $x = y$
EDIT: I just saw you couldn't use limits. I'll leave this answer up anyway ,as it might help others interested in the same question.
A: $z:=x-y;$ $x,y$ real.
$0 \le |z| \lt 1/n$ for $n \in \mathbb{Z^+}$.
Every $1/n$, $n \in \mathbb{Z^+}$, is an upper bound for $|z|$.
$A:= ${$1/n$, $n \in \mathbb{Z^+}$}. 
$\inf (A):= 0;$
$\rightarrow:$
$0 \le |z| \le \inf (A) =0.$
Hence $z = 0$.
We prove that $\inf (A)  =0.$
$0$ is a lower bound for $A$:
Assume there is a lower bound  $L$, real,  of $A$ with $L>0$, I.e
$0 \lt L \le 1/n , n \in \mathbb{Z^+}.$
There is a $ n_0  > 1/L .$
(Archimedes).
Then :  $0< 1/n_0 < L.$
Contradiction , 
hence $\inf(A) =0.$
