Does the above non-Archimedean but ordered field satisfy Nested interval property? Consider the ordered non-Archimedean field $ \mathbb{R}(t)$, the field of rational function.
My question is:
$ \text{Does the above non-Archimedean but ordered field satify Nested interval property?} $
Answer:
The field of rational function $\mathbb{R}(t)$ is expressed as
$ \mathbb{R}(t)=\left\{\frac{p(t)}{q(t)}: p(t), q(t) \in \mathbb{R}[t], \ q(t) \neq 0 \right\}$
This field is totally ordered field but non-Archimedean.
How to conclude about Nested interval property? 

A totally ordered field $\Bbb F$ is said to have the Nested interval property if every decreasing sequence of closed and bounded intervals in $\Bbb F$ has a nonempty intersection.

 A: Assume the ordering with $t>a$ for all $a \in \mathbb R$.  The ordering is:
$$
\frac{p(t)}{q(t)} > 0 \quad\Longleftrightarrow\quad \text { there exists } x_0 \in \mathbb R \text { with } \frac{p(x)}{q(x)}>0 \text { for all } x > x_0 .
$$
So my $t$ is $1/t$ in Chilote's answer.  
Use closed bounded intervals
$$
\left[n,\frac{1}{n}\;t\right]\qquad n=1,2,3,\dots
$$
We claim $Q=\varnothing$, where
$$
Q := \bigcap_{n=1}^\infty \left[n,\frac{1}{n}\;t\right] .
$$
First, if $\frac{p(t)}{q(t)} \le 0$ then $\frac{p(t)}{q(t)} \notin Q$.
Let $\frac{p(t)}{q(t)} > 0$.
Then $\frac{p(t)}{q(t)} \ge n$ for all $n$ implies $\deg p > \deg q$.
Also 
$\frac{p(t)}{q(t)} \le \frac{1}{n}\,t$ for all $n$ implies $\deg p \le \deg q$.
Thus $\frac{p(t)}{q(t)} \notin Q$.
A: 
$\mathbb{R}(t)$ does not satisfy the Nested interval property.

Proof 1:
Let's consider the order in $\mathbb{R}(t)$ such that $0<t<a$ for every $a\in\mathbb{R}^+$. Then the closed intervals $[-x^n,x^n]$ ($n\in\mathbb{N}$) form a nest with intersection $\{0\}$. Notice that the formal power series $f(t)=\sum_{i=1}^\infty t^{i^2}$ is not in $\mathbb{R}(t)$ but the partial sums $s_n(t)=\sum_{i=1}^n t^{i^2}\in \mathbb{R}(t)$ converges to $f(t)$ in $\mathbb{R}((t))$ (proofs of these statements here).
Now consider the nest formed by intervals of the form $I_n=s_n(t)+[-x^{n^2},x^{n^2}]$ for $n\in\mathbb{N}$. Notice that $I_{n+1}\subset I_n$.
If these intervals are considered in $\mathbb{R}((t))$ then its intersection will be $\{f(t)\}$. Therefore, if this nest is considered in $\mathbb{R}(t)$, then its intersection is empty.
Proof 2: When an ordered field satisfies the Nested interval property, then it is Cauchy-complete i.e. every Cauchy sequence is convergent in the order topology.
In fact, an ordered field is Cauchy complete whenever every nest of intervals with diameters tending to $0$ has nonempty intersection.
Notice that the field $\mathbb{R}((t))$ is the completion of $\mathbb{R}(t)$.
Since $\mathbb{R}(t)$ is not Cauchy complete,  it cannot satisfy the Nested interval property.
A: The nested interval property does not hold in $\mathbb{R}(t)$.  Here
is a proof, although it is probably not the simplest.
$\mathbb{R}(t)$ can be ordered in more than one way.  I will assume that
the ordering is the one where the polynomial
$a_dt^d + a_{d+1}t^{d+1} + a_{d+2}t^{d+2} + \cdots$
is positive iff $a_d > 0$.
(It seems more usual to take the ordering in which a polynomial is
positive iff its leading coefficient is positive; in that case, use
the infinitesimal element $1/t$ in place of $t$ in what follows.)
The element $u = 1 + t$ has no square root in $\mathbb{R}(t)$,
because if $(p(t)/q(t))^2 = 1 + t$, then $p(t)^2 = (1 + t)q(t)^2$,
which is impossible, because the highest power of $t$ on the left is
even and the highest power on the right is odd.
(In the case of the alternative ordering of $\mathbb{R}(t)$: if
$(p(t)/q(t))^2 = 1 + 1/t$, then $tp(t)^2 = (1 + t)q(t)^2$, which is
impossible, because the lowest power of $t$ on the left is odd and
the lowest power of $t$ on the right is even.)
Consider the fate of an iterative process which attempts to compute
a square root of $u$ in $\mathbb{R}(t)$.
Define a sequence
$(x_n)_{n\geqslant0}$ by $x_0 = 1$ and $x_{n+1} = f(x_n)$, where:
$$
f(x) = 1 + \frac{t}{1 + x} = \frac{u + x}{1 + x}.
$$
By induction on $n$, we have $x_n \geqslant 1$ for all $n$.
A simple calculation gives:
$$
f(x)^2 - u =
\frac{u^2 + 2ux + x^2 - u - 2ux - ux^2}{(1 + x)^2} =
-t\frac{x^2 - u}{(1 + x)^2},
$$
whence, by another induction on $n$:
$$
\left\lvert x_n^2 - u \right\rvert \leqslant \frac{t^{n+1}}{2^{2n}}
\quad (n = 0, 1, 2, \ldots).
$$
By another simple calculation:
$$
f(x) - x = \frac{u + x - x - x^2}{1 + x} = -\frac{x^2 - u}{1 + x},
$$
whence, by induction on $n$ again,
$$
\left\lvert x_{n+1} - x_n \right\rvert \leqslant
\frac{t^{n+1}}{2^{2n+1}}
\quad (n = 0, 1, 2, \ldots).
$$
As for the order properties of the sequence $(x_n)_{n\geqslant0}$,
first note that, by induction on $m$, using the results that have
already been proved,
$$
x_{2m}^2 < u < x_{2m+1}^2 \quad (m = 0, 1, 2, \ldots).
$$
Then, a third simple calculation:
$$
f^2(x) - x =
\frac{u + \frac{u + x}{1 + x}}{1 + \frac{u + x}{1 + x}} - x =
\frac{(u + 1)x + 2u}{2x + u + 1} - x =
-2\frac{x^2 - u}{2x + u + 1},
$$
followed by yet another inductive argument, gives:
$$
x_0 < x_2 < x_4 < \cdots < x_5 < x_3 < x_1.
$$
Suppose, if possible, that there exists $x$ such that:
$$
x \in \bigcap_{m=0}^\infty [x_{2m}, x_{2m+1}].
$$
Then, because $f$ is a decreasing function, we have:
$$
x_{2m+2} = f(x_{2m+1}) \leqslant f(x)
\leqslant f(x_{2m}) = x_{2m+1}
\quad (m = 0, 1, 2, \ldots).
$$
Therefore:
$$
\left\lvert f(x) - x \right\rvert \leqslant
\left\lvert x_{2m+1} - x_{2m} \right\rvert \leqslant
\frac{t^{2m+1}}{2^{4m+1}}
\quad (m = 0, 1, 2, \ldots).
$$
Let $\left\lvert f(x) - x \right\rvert = p(t)/q(t)$, where
$p(t), q(t) \in \mathbb{R}[t]$ and $q(t) \ne 0$. Then:
$$
p(t) \leqslant \frac{t^{2m+1}q(t)}{2^{4m+1}}
\quad (m = 0, 1, 2, \ldots).
$$
This is impossible for strictly positive $p(t)$, if $2m + 1$ exceeds
the lowest power of $t$ in $p(t)$. Therefore $p(t) = 0$; therefore
$f(x) = x$; therefore $x^2 = u$. But we saw that such
a value of $x$ does not exist in $\mathbb{R}(t)$, so we have a
contradiction, proving that the intersection of the nested sequence
of closed and bounded intervals $[x_{2m}, x_{2m+1}]$ is empty. $\square$
(In the case of the alternative ordering of $\mathbb{R}(t)$, we
have, instead:
$$
2^{4m+1}t^{2m+1}p(t) \leqslant q(t),
$$
which leads to a contradiction whenever $2m + 1$ exceeds the degree
of $q(t)$. $\square$)
