Solution:

We work with directed angles modulo $180^{\circ}$.
In this sketch, $T$ corresponds to the point $P$.

Step 1: First we note that $CA$ and $CB$ are the perpendicular bisectors of the segments $\overline{PB_1}$ and $\overline{PA_1}$, so $C$ is the circumcenter of the triangle $PA_1B_1$ and it follows that
$$\measuredangle A_1CB = \frac{1}{2} \measuredangle A_1CP = \measuredangle A_1B_1P = \measuredangle A_1B_1B_2 = \measuredangle A_1C_2B_2.$$

Step 2: Analogously to Step 1, it follows that $\measuredangle A_1BC = \measuredangle A_1B_2C_2$, so the triangles $A_1BC$ and $A_1B_2C_2$ are similar.

Step 3: Using the similarity just proved, it follows that
$$\measuredangle CA_1B = \measuredangle C_2A_1B_2 \quad \text{and} \quad \frac{\left|\overline{A_1C}\right|}{\left|\overline{A_1B}\right|} = \frac{\left|\overline{A_1C_2}\right|}{\left|\overline{A_1B_2}\right|}$$
and thus also
$$\measuredangle B_2A_1B = \measuredangle C_2A_1C \quad \text{and} \quad \frac{\left|\overline{A_1B_2}\right|}{\left|\overline{A_1B}\right|} = \frac{\left|\overline{A_1C_2}\right|}{\left|\overline{A_1C}\right|}$$
so the triangles $A_1B_2B$ and $A_1C_2C$ are also similar.

Step 4: Let $K$ be the point of intersection of $C_2C$ with $\Omega$. Then $180^{\circ} - \measuredangle A_1B_2K = \measuredangle A_1C_2K = \measuredangle A_1C_2C = \measuredangle A_1B_2B$ holds, so $K$ lies on the line $BB_2$. Analogously one can show that $K$ also lies on the line $AA_2$, which completes the proof.