Rudin Theorem $1.11$

Rewording it slightly: Every element of $B$ is an upper bound for $L$, so if $x\in B$ is less than $\alpha$, then $x$ is an upper bound for $L$ smaller than the least upper bound. This contradicts the definition of least upper bound, so no such $x$ exist. In other words, $\alpha\leq x$ for every $x\in B$.

I do not know why you included a restatement of the fact that $B$ is bounded below. The part you put in quotes follows directly from the last sentence of the previous paragraph in your post:

And because $\alpha = \sup L$, $\gamma < \alpha$ implies that $\gamma$ is not an upper bound of $L$ and $\gamma \notin B$ since every element of $B$ is an upper bound of $L$.

From here you could use contraposition:

$$\gamma<\alpha\implies \gamma\not\in B$$

is equivalent to

$$\gamma\in B \implies \gamma\geq \alpha.$$