Prove that if $a \ge c$ for all $c < b$, then $a \geq b$

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 $a \ge b$.