This works, right? Choose a random element $x_i$ from $S_i$ for $i=1,\cdots ,n$. The probability that the chosen elements from $S_i$ and $S_j$ are the same is $\tfrac{|S_i \cap S_j|}{|S_i||S_j|}$, so the probability that some two $x_i$ and $x_j$ are the same is at most $\textstyle\sum_{1\leq i<j\leq n}{\frac{|S_i \cap S_j|}{|S_i||S_j|}}. $ Since this is less than $1$, there is some choice for which all $x_i$'s are distinct. $\blacksquare$

Any solution without probability.

Consider the set $S$ of all ordered $n$-tuples $(x_1,x_2,\dots,x_n)\,,\, x_i\in S_i, i=1,2,\dots,n$. Obviously $|S|=|S_1|\cdot |S_2|\cdots|S_n|$. Let's estimate the number of "bad" tupples which contain non-distict elements. The number of tuples for which $x_i=x_j$ is $|S_i\cap S_j|\prod_{k\neq i,j}|S_k|={\frac{|S_i \cap S_j|}{|S_i|\cdot|S_j|}} \cdot |S|$ . Thus the number of bad pairs is at most: \[|S|\sum_{1\leq i<j\leq n}{\frac{|S_i \cap S_j|}{|S_i||S_j|}} < |S|\]which means not all tuples in $S$ are bad.

This is precisely https://artofproblemsolving.com/community/c6h225275p1252232 .

I don't think there is a solution without probability.

Yes, I did it with probab. too

ThE-dArK-lOrD wrote: Let $n$ be a positive integer, and let $S_1,S_2,…,S_n$ be a collection of finite non-empty sets such that $$\sum_{1\leq i<j\leq n}{\frac{|S_i \cap S_j|}{|S_i||S_j|}} <1.$$Prove that there exist pairwise distinct elements $x_1,x_2,…,x_n$ such that $x_i$ is a member of $S_i$ for each index $i$. This reminds me a lot of CGMO 2010/1.