By the given conditions we obtain
$$\sum_{i=1}^{n} \binom{|A_i|}{2} = \binom{n}{2}. \qquad (1)$$
Let $d_i = |\{k \mid i \in A_k\}|$. Clearly,
$$\sum_{i=1}^{n} d_i = \sum_{k=1}^{n} |A_k|. \qquad (2)$$
From the "existence and uniqueness" property, we obtain
$$\sum_{i=1}^{n} \binom{d_i}{2} = \sum_{1 \le i < j \le n} |A_i \cap A_j|.$$
Also, $|A_i \cap A_j| \le 1$ (otherwise the "uniqueness" property of the problem would be violated). To prove $A_i \cap A_j \ne \emptyset$ is equivalent to proving $|A_i \cap A_j| = 1$, which is equivalent to
$$\sum_{i=1}^{n} \binom{d_i}{2} = \binom{n}{2}.$$
By (1) and (2), together with the definition of the binomial coefficient $\binom{x}{2} = (x^2 - x)/2$, the above equality is in turn equivalent to
$$\sum_{i=1}^{n} d_i^2 = \sum_{k=1}^{n} |A_k|^2. \qquad (3)$$
Next, consider ordered pairs $(i, k)$ where $i \notin A_k$. Suppose $A_k = \{j_1, j_2, \dots, j_t\}$, then $\{i, y_1\}, \{i, y_2\}, \dots, \{i, y_t\}$ are $t$ distinct two-element subsets of some $A_j$ (since the two-element set $\{y_a, y_b\}$ is uniquely contained in $A_k$), so $d_i \ge |A_k|$, and we also have
$$\frac{d_i}{n - d_i} \ge \frac{|A_k|}{n - |A_k|}.$$

Summing the above inequality over all required $(i, k)$, we obtain
$$\begin{aligned} \sum_{i=1}^{n} d_i &= \sum_{i=1}^{n} \sum_{k|i \notin A_k} \frac{d_i}{n - d_i} \\ &\geq \sum_{i=1}^{n} \sum_{k|i \notin A_k} \frac{|A_k|}{n - |A_k|} \\ &= \sum_{k=1}^{n} \sum_{i|i \notin A_k} \frac{|A_k|}{n - |A_k|} \\ &= \sum_{k=1}^{n} |A_k|. \end{aligned}$$
By equality (2), the above must be an equality, and finally we get $d_i = |A_k|$ when $i \notin A_k$. Next, consider the equality
$$\begin{aligned} \sum_{i=1}^{n} (n - d_i) d_i &= \sum_{i=1}^{n} \sum_{k|i \notin A_k} d_i \\ &= \sum_{i=1}^{n} \sum_{k|i \notin A_k} |A_k| \\ &= \sum_{k=1}^{n} \sum_{i|i \notin A_k} |A_k| \\ &= \sum_{k=1}^{n} (n - |A_k|) |A_k|. \end{aligned}$$
Combined with equality (2), this yields (3). This completes the proof!