Prove: If $\displaystyle S \subset \bold{R} $, then $\displaystyle \bar{S} $ is the smallest closed set containing $\displaystyle S $.

So $\displaystyle \bar{S} = S \cup S' $. We want to show that if $\displaystyle T $ is any closed set and $\displaystyle S \subset T $, then $\displaystyle \bar{S} \subset T $.

So do proof by contradiction: Suppose that $\displaystyle \bar{S} $ is not the smallest closed set containing $\displaystyle S $. Is this the correct way to proceed?