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.

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.
