Nested Interval Theorem; free of set theory I am reading this proof on Spivak for the Nested Interval Theorem. I pretty much did exactly what he did except the last step.

For each $m$ and $n$ we have $a_n \leq b_m$, because $a_n \leq a_{n+m} \leq b_{n+m} \leq b_m$. It follows from problem 12 that $\sup \{a_n : n \in \mathbb{N} \} \leq \inf \{b_n : n \in \mathbb{N} \}$. Let $x$ be any number between these two numbers. Then $a_n \leq x \leq b_n$ for all $n$, so $x$ is in every $I_n$

Now could someone please explain to me how is it that he can simply "let $x$ be any number between these two numbers"? My original goal was to show that the sup and inf are indeed the same thing and therefore we may choose this number to be our $x$. 
 A: What is there to explain? Let $u=\sup\{a_n:n\in\Bbb N\}$ and $v=\inf\{b_n:n\in\Bbb N\}$. From a previous result you know that $u\le v$. Therefore the interval $[u,v]$ is non-empty, and we can choose an arbitrary $x\in[u,v]$, i.e., an $x\in\Bbb R$ such that $u\le x\le v$. Since $a_n\le u\le x\le v\le b_n$ for each $n\in\Bbb N$, it is certainly true that $x\in[a_n,b_n]$ for each $n\in\Bbb N$ and hence that
$$x\in\bigcap_{n\in\Bbb N}[a_n,b_n]\;,$$
which is all that is being asserted at that point. Now it may be that you have other information that will enable you to show that in fact $u=v$, so that $[u,v]=\{u\}=\{v\}$, and therefore $x=u=v$, but that’s a separate issue requiring a separate proof. At this point we know only that the intervals have non-empty intersection, not that the intersection is a singleton.
A: If the hypotheses of this Nested Interval Theorem are the usual ones, namely that each interval is nonempty and closed and includes the next interval in the sequence, your "original goal" of showing the sup and the inf are the same will be unattainable, because that conclusion might simply be false.  Suppose, for example, that $a_n=1-\frac1n$ and $b_n=7+\frac1n$.
