I am thinking, primarily, in terms of $\mathbb{Q}$ and $\mathbb{R}$, but it would be nice to see if this also holds generally.
In Chapter 1 of Rudin, he asserts that $\mathbb{Q}$ is a subfield of the ordered field $\mathbb{R}$. $\mathbb{Q}$ is itself an ordered field, but this brings me to two questions.
First, is every subfield of $\mathbb{R}$ necessarily an ordered field? Does the subfield necessarily inherit the order from the ambient field? It makes sense to me that the subfield would necessarily inherit the operations from the ambient field (in the same way that a subgroup inherits its binary operations from the ambient group), but is this also the case for an order?
First, it is trivial but important to notice that any subset $S$ of an ordered set $T$ inherits the order from $T$: if $x,y\in S$, then since we can compare them in $T$, we can certainly compare them in $S$; the ambient set does not matter. Of course $T$ has many other orders (if it is finite with $n$ elements, it has $n!$ total orders, for instance), but only one comes from the specific order we consider on $S$. (Of course, $S$ also has many orders, all of them induce an order on $T$, usually different, but sometimes different orders on $S$ may induce the same order on $T$).
Now when we talk about ordered fields, we don't want to include any orders; just those which are compatible with sums and products. In general, a field may have any number of compatible orders: $0$ (like $\mathbb{C}$), $1$ (like $\mathbb{R}$ or $\mathbb{Q}$), finitely many (like $\mathbb{Q}[\sqrt{2}]$), or infinitely many (like $\mathbb{Q}(X)$). A simple result is that a field admits at least one order if and only if $-1$ is not a sum of squares.
It is easy to see that restricting a field order to a subfield $K\subset L$ still gives a field order: since the operations on $K$ are the same as those on $L$, and since the order is compatible with the operations on $L$, it is compatible with the operations on $K$.
So any subfield of $\mathbb{R}$ has a natural field order. But it may also have different field orders that are not the restriction of the field order on $\mathbb{R}$ (which on the other hand, is unique because any real number is a square or the opposite of a square, so its sign is determined by the algebra).
For instance, when we write $K=\mathbb{Q}[\sqrt{2}]$, we mean it as a subfield of $\mathbb{R}$, so we have a natural order on it, for which $\sqrt{2}>0$. But from a purely algebraic point of view, $K$ has two square roots of $2$, and they are completely indistinguishable (they are conjugate under the Galois action). So actually we could just as well decide that the element we called "$\sqrt{2}$" is negative. This gives an other order on $K$, which also corresponds to the other embedding $K\to \mathbb{R}$ (if $x\in K$ is one of the two square roots of $2$, we can send $x$ either to $\sqrt{2}$ or $-\sqrt{2}$).