Prove that if $a \ge c$ for all $c < b$, then $a \geq b$ 
Let $a$ and $b$ be elements in an ordered field, prove that if $a \ge c$ for every $c$ such that $c \lt b$, then $a\ge b$.

My proof idea below:
Let $S = \{x | x<b\}$. Then $a$ is an upper bound for $S$. If I can show that $b$ is the least upper bound for $S$, then it follows from the definition of least upper bound that $a\ge b$.
However, I have a hard time proving the claim that $b$ is the least upper bound for $S$. Am I on the right direction? Can anyone help? Thank you.
 A: Suppose
$a < b$.
Let $d = b-a$,
so that $d > 0$.
Let $c = b-d/2$.
Then $c < b$,
but 
$c 
= b-d/2
= b-d+d/2
=a+d/2
> a
$
which contradicts the assumption
that $a \ge c$
for every $c < b$.
Therefore
$a \ge b$.
A: This post is intended to point out a flaw in an earlier version of the question by providing a counter-example to an impossible proof.
Let $a = b^-$, where $b^- = c$ and c is the surreal number infinitely close to b such that no number is between c and b and c < b. Then a < b. Conjecture contradicted. Please think more carefully next time.
A: As noted above, $a$ must be greater than $c$ for this to possibly be provable. Possibly check with your instructor? Typos do happen...
A: Let $S=\{c:c<b\} .$ By hypothesis, $\forall c\in  S\;(c<a).$ So if $a\in S$ then $a<a,$ which cannot be, because "$<$" is irreflexive. The whole field is equal to $S\cup \{b\} \cup \{d:d>b\} $ because  "$<$" satisfies trichotomy. Since $a\not \in S $ we have $a\in \{b\}\cup \{d:d>b\}.$ QED. 
Note that this applies to any linearly ordered set.
A: think about the definition of a ordered field. Since a and b are elements of the field, (b-a) is in F. thus (b-a)'s multiplicative inverse is in F. call this inverse x. we get x(b-a) = e. xb-xa=e. note that e is not the smallest elements of the ordered field, why? pick y>e, then y inverse must be smaller than e (this may not be true in a general field, but should be true in an ordered field, by the properties of the ordered field.) then from here, you can construct an element between a and b and the conclusion will follow. This is what i would do. but not 100% sure tho. 
