The Neyman–Pearson lemma

Step 1 (set up an arbitrary competitor test): let $C$ be the likelihood-ratio rejection region from the statement, with indicator function $\phi_C$, and let $D$ be the rejection region of any other test with $P(D \mid H_0) \leq \alpha$, with indicator function $\phi_D$. We must show the power at $\theta_1$ satisfies $P(C \mid H_1) \geq P(D \mid H_1)$.

Step 2 (a key pointwise inequality): for every outcome $x$, the quantity $(\phi_C(x) - \phi_D(x)) \cdot (L(\theta_1) - k \, L(\theta_0))$ is never negative. Indeed, if $x \in C$ then $L(\theta_1) > k \, L(\theta_0)$ and $\phi_C(x) - \phi_D(x) \geq 0$ since $\phi_C(x) = 1$; if $x \notin C$ then $L(\theta_1) \leq k \, L(\theta_0)$ and $\phi_C(x) - \phi_D(x) \leq 0$ since $\phi_C(x) = 0$; in both cases the product of the two factors is $\geq 0$.

Step 3 (integrate the inequality): summing (or integrating) this nonnegative quantity over all outcomes $x$ gives $\sum_x (\phi_C(x) - \phi_D(x)) \, L(\theta_1) - k \sum_x (\phi_C(x) - \phi_D(x)) \, L(\theta_0) \geq 0$, which rewrites as $\big(P(C \mid H_1) - P(D \mid H_1)\big) - k \big(P(C \mid H_0) - P(D \mid H_0)\big) \geq 0$.

Step 4 (use the significance levels to conclude): by construction $P(C \mid H_0) = \alpha$, and by assumption $P(D \mid H_0) \leq \alpha$, so $P(C \mid H_0) - P(D \mid H_0) \geq 0$; since $k \geq 0$, subtracting $k$ times a nonnegative quantity from Step 3's first bracket only strengthens the inequality $P(C \mid H_1) - P(D \mid H_1) \geq k \big(P(C \mid H_0) - P(D \mid H_0)\big) \geq 0$, so $P(C \mid H_1) \geq P(D \mid H_1)$, as required.