Let $B \subseteq S$ with $\inf(B) \in S.$ Then $\inf(B) = \sup\{\text{lower bounds of }B\}$.
Proof: Let $L = \{\text{lower bounds of }B\}$ and $\alpha = \sup L.$ Then for any $\gamma \in L, \exists \gamma_1 \in L$ s.t. $\gamma < \gamma_1 < \alpha$ meaning $\gamma \not \in B^{1}.$ Thus $\alpha \le^{2} x \in B$ and so $\alpha$ is a lower bound of $B$. Any $\beta > \alpha $ cannot be in $L$ as $\alpha$ is an upper bound meaning $\alpha = \inf(B).$
Questions.
$^1$ Why is $\gamma \not \in B$? Is it because if $\gamma \in B$, then $\gamma_1$ stops being a lower bound as there's a smaller element in $B$, namely, $\gamma$?
$^2$ Why do need to show no element of $L$ less than $\alpha$ is in $B$ to conclude $\alpha \le x \in B$ for all $x$? How does this conclusion follow?
I found this proof in Rudin, $3$rd edition (p $5$, theorem $1.11$). Reading it as written by Rudin seems to make it clearer. I'll answer my on question so that I do not forget what made me understand it.
$^1.$ Every $x \in B$ is an upper bound of $L$. If $\gamma < \alpha,$ then $\gamma$ is not an upper bound of $L$, so $\gamma$ cannot be in $B$.
$^2.$ We know $\gamma < \alpha \implies \gamma \not \in B.$ That means no element less than $\alpha$ is in $B$ implying all the elements of $B$ must be greater than or equal to $\alpha.$