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.

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. 
