Let the elements of $X$ be $a_1<a_2<...<a_8$ and denote the seven successive differences by $d_i=a_{i+1}-a_i.$
Consider the subsets of size $4$ which contain either $2$ or $3$ elements of $\{a_5,a_6,a_7,a_8\}$. There are $$\begin{pmatrix}4\\1\\\end{pmatrix}\begin{pmatrix}4\\3\\\end{pmatrix}+\begin{pmatrix}4\\2\\\end{pmatrix}\begin{pmatrix}4\\2\\\end{pmatrix}=52$$ of these subsets and the possible sums of their elements range from $a_1+a_2+a_5+a_6$ to $a_4+a_6+a_7+a_8$. So, by the pigeon-hole principle, we are finished unless $$a_4+a_6+a_7+a_8-(a_1+a_2+a_5+a_6)+1\ge 52$$ $$\text {i.e.} 2(a_8-a_1)\ge51+d_1+d_4+d_7.$$
Since $a_8-a_1\le 29$ we must have $d_1+d_4+d_7\le7$. Using the observations given below, $d_1,d_4,d_7$ are all different and no two can add to the third and so $\{d_1,d_4,d_7\}=\{1,2,4\}$ and $\{a_1,a_{8}\}=\{1,30\}.$
Some observations about the $d_i$.
(1) Any two non-adjacent differences are unequal.
(2) Given three non-adjacent differences, none is the sum of the other two.
(3) Given two adjacent differences, the sum of these differences can replace one of the differences in observations (1) and (2). (We still require the 'combined difference' to be non-adjacent to the other differences involved.)
The proofs of these are all elementary and of the same form. As an example, suppose we have $d_2+d_3=d_5+d_7$, which is a combination of (2) and (3). Then $$a_4-a_2=a_6-a_5+a_8-a_7.$$
The sets $\{a_4,a_5,a_7\}$ and $\{a_2,a_6,a_8\}$ then have the same sum and $a_1$, say, can be added to each.
To return to the main proof where we know that the differences $\{d_1,d_4,d_7\}=\{1,2,4\}$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $1$. Then, by the observations, $\{d,d+1\}\cap\{2,4,6\}$ is empty. So $d\ge7$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $2$. Then, by the observations, $\{d,d+2\}\cap\{1,3,4,5\}$ is empty. So $d\ge6$.
Let $d$ be a difference adjacent to whichever of $\{d_1,d_4,d_7\}$ is $4$. Then, again by the observations, $\{d\}\cap\{1,2,3\}$ is empty. So $d\ge4$.
The sum of the differences (which is $29$) is now at least $(1+2+4)+(7+6+4)+d$, where $d$ is the 'other' difference adjacent to $d_4$. Therefore $d_4=4$ and the two differences adjacent to it (which cannot be equal) are $4$ and $5$. The differences adjacent to the differences of $1$ and $2$ are thus forced to be $7$ and $6$, respectively. Then $a_1+a_8=a_3+a_5$ and we are finished.
Your attempt 1 is actually a correct approach.
Assume that for an odd integer $k$ not divisible by 3, box $k$ contains all the odd numbers from $\{1,2,\ldots, 3n\}$ that are of the form $3^ik$, where $i$ is a non-negative integer. Then among any two numbers in a box $k$, one divides the other.
For example box 1 contains all the numbers of form $3^i$ and box 5 all the numbers of form $5\cdot 3^i$
Now, you have box 1, box 5, box 7, box 11, box 13, box 17, etc, so the number of boxes is the number of odd integers in $\{1,\ldots, 3n\}$ that are not divisible by 3.
In case when $n$ is even, the number of odd integers is $3n/2$ and $n/2$ of them are divisible by 3, so the number of boxes is $n$.
In case when $n$ is odd the number of odd integers is $(3n+1)/2$ and $(n+1)/2$ of them are divisible by 3, so the number of boxes is again $n$.
So among your $n+1$ numbers at least two belong to the same box, say box $k$. Then one of the numbers is $3^ik$, another is $3^jk$, so clearly one divides another.
Best Answer
The answer is 16: to show that 15 is possible, just take $2,3,5,7,\ldots,47$.
Now for the fun part, making the pigeonholes. The first pigeonhole is $$S_1=\{2,4,6,8,\ldots\}$$ and the second is $$S_2=\{3,9,15,\ldots\}$$ Continuing thus, the $i$th set is (for $i>1$) $$S_i=\{p_i,\ldots\}$$ where $p_i$ is the $i$th prime, and the elements that occur in the set are multiples of $p_i$ which don't occur in previous sets.
Note that every positive integer ends up in one of these sets; in particular, the numbers 2-50 end up in the first 15 of these sets. Thus taking $16$ numbers guarantees two in the same set (by PHP) and we're done, since any two numbers in a set trivially share a factor.