Mistake in a proof of termination phase of Simplex algorithm in CLRS?
Because of line 9, $\Delta_l$ is going to be $\infty$ if and only if all the $\Delta_i$ with $i \in B$ are $\infty$. This can happen only when all the $a_{ie}$ with $i \in B$ are $ \le 0$. Hence, by construction the solution $\bar{x}$ satisfies $\bar{x}_i \ge 0$ for all $i$. Therefore it is both feasible and it has unbounded objective value. 

If you are confused by the $\infty$, as Marcus Ritt mentioned in a comment you can replace the $\infty$ in the definition of $\bar{x}$ with a very large value $M$ and consider what happens as $M$ tends to $\infty$.