Monotone Convergence Property $iff$ Order Completeness in an arbitrary ordered field.

By induction on $n$. Suppose that $U-epsilonsubseteq U$. If $U-nepsilonsubseteq U$ for some $ninBbb N$, then

$$U-(n+1)epsilon=(U-epsilon)-nepsilonsubseteq U-nepsilonsubseteq U,.$$

That’s the induction step, and the claim follows immediately.

Added to answer the additional questions.

For the first question, note that $U-frac1nsubseteq U-frac1m$ whenever $nge m$.¹ Thus, if $xnotin U-frac1m$, then $xnotin U-frac1n$ for $nge m$. There is certainly some $m$ such that $x_2notin U-frac1m$, and we’ve just shown that in that case $x_2notin U-frac1n$ for all $nge m$, so we can simply take $$n_2=max{m,n_1+1}$$ to ensure that $n_2>n_1$.

¹ To see this, suppose that $nge m$, and let $xin U-frac1n$. Then $$x+frac1mge x+frac1nin U,,$$ so $x+frac1min U$, and hence $xin U-frac1m$.

For your second question, note that if $uin U$, and $v>u$, then $vin U$. If the sequence is not bounded above by each element of $U$, there are a $kinBbb Z^+$ and a $uin U$, such that $u<x_k$. But then $x_kin U$, contradicting the fact that every point of the sequence was chosen not to be in $U$.

For your third question, the fact that

$$bigcap_{j=1}^inftyleft(U-frac1{n_j}right)=bigcap_{n=1}^inftyleft(U-frac1nright)$$

is an immediate consequence of the fact, proved above, that $$U-frac1nsubseteq U-frac1m$$ whenever $nge m$:

$$U-1supseteq U-frac12supseteq U-frac13supseteqldots;.$$

If a point is not in some $U-frac1n$, then it isn’t in $U-frac1{n_j}$ for any $n_jge n$, and there is certainly an $n_jge n$. (In fact the intersection of any infinite subsequence of a decreasing sequence of sets is the same as the intersection of the entire sequence.)