What makes this a $1- \alpha$ confidence set? From Statistical Inference by Casella and Berger:

Let $T$ be a discrete statistic with cdf $F_T(t|\theta) = P(T \le t | \theta).$  Let $\alpha_1 + \alpha_2 = \alpha$ with $0 \lt \alpha \lt 1$ be fixed values. Suppose that for each $t \in \tilde T$, $\theta_L(t)$ and $\theta_U(t)$ can be defined as follows:
If $F_T(t | \theta)$ is a decreasing function of $\theta$ for each $t$, define $\theta_L(t)$ and $\theta_U(t)$ by
$$P(T \le t | \theta_U(t)) = \alpha_1 \text{ and } P(T \ge t | \theta_L(t)) = \alpha_2$$
Since $P_{\theta}(F_T(T |\theta) \le x) \le x$ and this property holds for $\bar F_T(t|\theta) = P(T \ge t | \theta)$, this implies that the set $$\{\theta : F_T(T | \theta) \le \alpha_1 \text{ and } \bar F_T(T | \theta) \le \alpha_2\}$$
is a $1 - \alpha$ confidence set.

How do these assumptions show that this set is a $1-\alpha$ confidence set?  I can't figure out a way to show this.
 A: I believe there is a typo. As evidence, have a look at the following notes, which are (as far as I can tell) just a copy-paste of this section of Casella and Berger, except that the part that you're quoting from has been modified, to use a different confidence interval, on p. 24: http://www.stat.purdue.edu/~fmliang/STAT611/st611lect9.pdf
We can also try and see why this must be a typo directly. From the fact that $P_\theta (F_T (T \mid \theta) \leq x)\leq x$, we see in particular that $P_\theta (F_T (T \mid \theta) \leq \alpha_1)\leq \alpha_1$. Likewise, $P_\theta (\bar{F}_T (T \mid \theta) \leq \alpha_2)\leq \alpha_2$. Thus $$P_\theta [(F_T (T \mid \theta) \leq \alpha_1 )\vee (\bar{F}_T (T \mid \theta) \leq \alpha_2)]\leq\alpha$$
In other words, $$P_\theta [(F_T(T\mid \theta)\geq \alpha_1 )\wedge(\bar{F}_T(T\mid\theta)\geq \alpha_2)]\geq1-\alpha.$$
In other words, for this specific $\theta_0$, the test indicated in the brackets above is a level $\alpha$ test. Therefore, the inversion theorem between acceptance regions and confidence sets (Thm. 9.2.2 in Casella and Berger) implies that $$\{\theta : F_T(T\mid\theta)\geq \alpha_1 \wedge \bar{F}_T (T\mid \theta)\geq \alpha_2\}$$ is a $1-\alpha$ confidence set. 
Note: if you read further in the section of Casella and Berger you quote from (i.e. the next paragraph of Theorem 9.2.14), you can also see that they implicitly make another sign error: if taken literally, the set that your quote claims is the confidence interval should actually equal ${\theta : \theta > \theta_U, \theta < \theta_L}$, which is the complement of the intended interval!
