Let $(F,+,\cdot,0_F,1_F,\le)$ be an ordered field. If the set $\{x\in F: 0_F\le x\le 1_F\}$ is complete in the sense that the least upper bound property is satisfied, is the whole field complete, in the sense that the least upper bound property is satisfied?
Edit:
With DanielWainfleet's help, here's my attempt to proof the statement.
For any ordered field $(F,+,\cdot,0_F,1_F,\le)$, $A=\{1_F,1_F+1_F,1_F+1_F+1_F,\ldots\}\subseteq F$. Since $0_F< 1_F$, $1_F<1_F+1_F<1_F+1_F+1_F<\ldots$. So, all members of $A$ are distinct. Let $2_F=1_F+1_F$, $3_F=1_F+1_F+1_F$, etc. That is, $A=\{1_F,2_F,3_F,\ldots\}$. There is an isomorphism from $A$ to $\mathbb{N}$, i.e. $f:A\rightarrow\mathbb{N}$ such that $f(x_F)=x$.
Because $(F,+,\cdot,0_F,1_F,\le)$ is a field, there is an additive inverse for every member of $A$. Let $-x_F$ be the additive inverse for any $x_F\in A$. Define $B=A\cup\{-x_F:x_F\in A\}\cup\{0_F\}$. There is an isomorphism from $B$ to $\mathbb{Z}$. (Such an isomorphism is obvious and hence will not be specified).
Because $(F,+,\cdot,0_F,1_F,\le)$ is a field, there is a multiplicative inverse for every member of $B$. Let $\frac{1_F}{y_F}$ be the multiplicative inverse for every member $y_F\in B$ such that $y_F\ne 0_F$. Since $(F,+,\cdot,0_F,1_F,\le)$ is a field, there is a member of $F$ that is equal to $x_F\cdot \frac{1_F}{y_F}=\frac{x_F}{y_F}$ for every $x_F\in B$. Define $C=B\cup\{\frac{x_F}{y_F}:x_F\in B \land y_F\in B-\{0_F\}\}$. There is an isomorphism from $C$ to $\mathbb{Q}$. (Such an isomorphism is obvious and hence will not be specified).
Because the set $\{x\in F: 0_F\le x\le 1_F\}$ satisfies the least upper bound property, every nonempty subset of it that has an upper bound has a least upper bound. Since the least upper bound property is equivalent to the greatest lower bound property, every nonempty subset of it with a lower bound has a greatest lower bound.
Suppose $S=\{\frac{1}{x_F}:x_F\in A\}$. The greatest lower bound for $S$ is $0_F$. The proof is given in DanielWainfleet's comment. This shows that there are no positive infinitesimals in $\{x\in F: 0_F\le x\le 1_F\}$, i.e. there is no $y\in\{x\in F: 0_F\le x\le 1_F\}$ such that $0_F<y<\frac{1}{x_F}$ for all $x_F\in A$.
Now, take an arbitrary $x\in F$ and consider the following set $\{q\in C:q<x\}$. For reductio, assume that $x$ is not the least upper bound for $\{q\in C:q<x\}$. By the definition of $\{q\in C:q<x\}$, $x$ is an upper bound for it. Let $y\in F$ be an upper bound for $\{q\in C:q<x\}$ such that $y<x$. Re-arranging, $0<x-y$. Find an $z\in A$ such that $\frac{1}{z}<x-y$. Such an $z$ exists because there are no positive infinitesimals. Now, find a $k\in B$ such that $\frac{k}{z}\le y<\frac{k+1}{z}$. Then, $\frac{k+1}{z}<x$. For if $\frac{k+1}{z}\ge x$, then $y<\frac{k}{z}$, which is absurd. So, we have $y<\frac{k+1}{z}<x$. Because $\frac{k+1}{z}\in C$ and $\frac{k+1}{z}<x$, $\frac{k+1}{z}\in\{q\in C:q<x\}$. But this implies that $y$ is not an upper bound for $\{q\in C:q<x\}$. Contradiction. So, there is no upper bound for $\{q\in C:q<x\}$ that is smaller than $x$. Therefore, $x$ is the least upper bound for $\{q\in C:q<x\}$.
With this, we can prove that $F$ is isomorphic to $\mathbb{R}$. Let $g:F\rightarrow \mathbb{R}$ such that $g(x)$ is the least upper bound for the set $\{q\in \mathbb{Q}:q<g(x)\}$. Since $F$ is isomorphic to $\mathbb{R}$, $(F,+,\cdot,0_F,1_F,\le)$ is complete, in the sense that the least upper bound property is satisfied.
In an ordered field, any two intervals $[a,b]$ and $[c, d]$ are order isomorphic ( by a suitable affine map $t\mapsto \alpha t + \beta$). So your statement is true.
Moreover:
(like @Daniel Wainfleet mentioned) every complete ordered field is order isomorphic to $\mathbb{R}.
Every archimedean ordered field is isomorphic to a unique subfield of $\mathbb{R}$. Every automorphism of such a field is the identity.
There exist ordered fields that are not complete, yet in some sense their algebra is similar to $\mathbb{R}$. For instance, consider the field consisting of all real algebraic numbers.