Assume there are no two such elements. Let $A_1, \ldots, A_{22}$ be the initial sets, $|A_i| = 5$, $i = 1, \ldots, 22$.
Put $S = \bigcup_{i=1}^{22} A_i$ and $a(x, y) = |\{i \mid \{x, y\} \subset A_i\}|$ for $x, y \in S$.
Assume there are five sets that contain $x$ and $y$, without loss of generality, $A_1, A_2, A_3, A_4, A_5$. Then a set $A_j$ with $\{x, y\} \not\subset A_j$ must contain an element from $A_i \setminus \{x, y\}$ for $i = 1, 2, 3, 4, 5$. Hence, $\{1, 2\} \cap A_j = \emptyset$ and a contradiction $|A_1 \cap A_j| = 2$ follows.
Consequently, we can assume that $a(x, y) \le 4$ for all $x, y \in S$.
There are exactly 10 unordered pairs contained in $A_1$, and every $A_i$, $i = 2, 3, \ldots, 22$, contains exactly one of them. Hence, one of the pairs is contained in $A_1$ and three other sets, i.e. there are $x, y \in S$ with $a(x, y) = 4$.
Without loss of generality, we assume that $x = 1, y = 2$ and that the four sets containing 1 and 2 are:
$$A_1 = \{1, 2, 3, a, b\}, \quad A_2 = \{1, 2, 4, c, d\}, \quad A_3 = \{1, 2, 5, e, f\}, \quad A_4 = \{1, 2, 6, g, h\}.$$
Furthermore, we can assume that $A_5 = \{1, 3, 4, 5, 6\}$. To meet the condition $|A_i \cap A_j| = 2$ for $i = 1, 2, 3, 4$ every $A_j$ with $j > 4$ must contain either 1 or 2. Assume that $1 \in A_j$ for $j = 5, 6, \ldots, 14$. Each of the nine sets $A_j$, $j = 6, \ldots, 14$, must contain $3, 4, 5$, or 6 because of $|A_5 \cap A_j| = 2$. Hence, there is an $x \in \{3, 4, 5, 6\}$ with $a(1, x) \ge 5$, a contradiction. Consequently, at most nine of the sets $A_j$, $j = 5, \ldots, 22$, contain 1. Analogously, at most nine of the sets $A_j$, $j = 5, \ldots, 22$, contain 2. It follows that exactly nine of the sets $A_j$, $j = 5, \ldots, 22$, contain 1, without loss of generality, the sets $A_5, \ldots, A_{13}$.
Taking into account that $a(x, y) \le 4$ for all $x, y \in S$ and $|A_i \cap A_j| = 2$ for $1 \le i < j \le 13$, without loss of generality, we can assume
$$A_6 = \{1, 3, c, e, g\}, \quad A_7 = \{1, 3, d, f, h\}, \quad A_8 = \{1, 4, a, e, h\},$$
$$A_9 = \{1, 4, b, f, g\}, \quad A_{10} = \{1, 5, a, d, g\}, \quad A_{11} = \{1, 5, b, c, h\},$$
and there is no proper choice left for $A_{12}$.