First, we select all the 3-element subsets of $X_1 = \{1, 2, 3, 4, 5\}$ and of $X_2 = \{6, 7, 8, 9, 10\}$. Then, the number of subsets will be $\binom{5}{3} + \binom{5}{3} = 20$. Suppose there is a covering set of size $k \le 5$, then one of the two sets $X_1, X_2$ will have no more than 2 elements, which means that there are 3 elements belonging to either $X_1$ or $X_2$ that are not selected. Clearly, the given condition is not satisfied. Thus, $k \ge 6$.

We will prove that there always exists a covering set of size 6. We will show that there are 4 numbers $a, b, c, d \in X$ of which at most 2 of these 4 numbers are selected into the given subsets. Suppose by contradiction that for every set of 4 numbers taken from $X$, there is always a subset of them chosen. We count in two ways the number of connections between the 4 sets of numbers $(a, b, c, d)$ taken from $X$ and the given subsets.

* Counting by tuple $(a, b, c, d)$: there are at least $\binom{10}{4} = 210$ connections.
* Counting by subsets: each subset of 3 elements will belong to 7 different tuples of 4 numbers, there are 20 subsets are given, so the number of connections is $7 \cdot 20 = 140 < 210$.

From here we see that there is an absurdity, so there should always be 4 numbers $a, b, c, d$ as stated. At this point, we choose the remaining 6 numbers in $X$. Then, each subset of 3 elements in the 20 given sets will always have at least one number belonging to these 6 values. So the minimum value to find is $k_{\min} = 6$. $\square$