Let $A$ be a countable subset of $\mathbb R$ which is well ordered with respect to usual ordering $\leq$ of $\mathbb R$.Then does $A$ have an order preserving bijection with a subset of $\mathbb N$?
Solution: No,the statement is not true.Take the example $A=\{1-\frac{1}{n}:n\in \mathbb N\}\cup \{1\}$.Then take any non-empty subset $B$ of $A$.Let $P=\{p\in \mathbb N:1-\frac{1}{p}\in B\}$.If $P=\phi$ then $B=\{1\}$ as $B\neq \phi$.If $P\neq \phi$,then by well ordering property of $\mathbb N$,there is a least element of $P$,call it $p_0$.Then $p_0\leq p$ for all $p\in P$.So,$1-\frac{1}{p_0}\leq 1-\frac{1}{p}$ for all $p\in P$.Now, for any $b\in B$ such that $b\neq 1$,$b=1-\frac{1}{k}$ for some $k$.
But,then $k\in P$ and hence $1-\frac{1}{p_0}\leq 1-\frac{1}{k}$ and $1-\frac{1}{p_0}\in B$,so it is the least element of $B$.So,$A$ is well-ordered.But,there is no order isomorphism of $A$ with $\mathbb N$.Is the counterexample correct?
I think your counterexample has a typo: the set $A$ you defined does not include $1$! You really want $A = \{1 - 1/n : n \in \mathbb{N}\} \cup \{1\}$ (assuming $\mathbb{N} = \{1, 2, \dots\}$).
Otherwise, everything is technically correct, but I have a couple small comments on style if you'd like to hear them:
Just to be precise, you should handle the case $b = 1$ when proving $1 - 1/p_0$ is the minimum element of $B$.
It would be good to explain in your proof why $A$ does not have any order isomorphism with $\mathbb{N}$.
Good work!