Union of sets having $\binom{n - 1}{2}$ elements with each two of them having $n - 2$ common elements has cardinality of at least $\binom{n}{3}$. 
Let $S_1, S_2, \dots, S_n$ sets that have each of them $n - 1 \choose 2$ elements, with $n - 2$ common elements for each two of them. Prove that their union has at least $n \choose 3$ elements. Find an example for equality case.

Let us examine the simplest case $n = 3$. Each of them has $2 \choose 2$ $= 1$ elements and each two of them have $1$ element in common, so, in fact, the three sets are identical, so their union has exactly $3 \choose 3$ $= 1$. (Equality case confirmed)
Examining $n = 4$, each set should have $3 \choose 2$ $= 3$ elements and $2$ in common every $2$. We may construct $\{a, b, c\}$, $\{a, b, d\}$, and the last set may be $\{a, b, e\}$ or $\{a, c, d\}$, both of them satsifying the conditions.
However, I am unable to generalize the problem

Here is an alternative approach based on @BillyJoe's formulation.  Let $x_j$ be the number of elements that appear in exactly $j$ sets.  Then the two conditions in the problem imply the following two linear equality constraints:
\begin{align}
\sum_{j=1}^n j x_j &= n\binom{n-1}{2} = 3\binom{n}{3} \tag1 \\
\sum_{j=1}^n \binom{j}{2} x_j &= (n-2)\binom{n}{2} = 3\binom{n}{3} \tag2
\end{align}
The total number of distinct elements is $\sum_{j=1}^n x_j$, which we want to show is at least $\binom{n}{3}$.
First note that $(j-2)(j-3) \ge 0$, which is true for all integer $j$, is equivalent to
$$1\ge \frac{2}{3}\cdot j-\frac{1}{3}\binom{j}{2}.\tag3$$
Now multiply $(1)$ by $2/3$ and $(2)$ by $-1/3$ and add them up to obtain
$$\sum_{j=1}^n x_j \stackrel{(3)}{\ge}
\frac{2}{3}\sum_{j=1}^n j x_j
-\frac{1}{3}\sum_{j=1}^n \binom{j}{2} x_j 
\stackrel{(1),(2)}{=} \frac{2}{3}\cdot3\binom{n}{3}
-\frac{1}{3}\cdot3\binom{n}{3}
= \binom{n}{3}.
$$
This lower bound is attained by taking $x_3=\binom{n}{3}$ and all other $x_j=0$.
