Let's play a game:
First, you pick five numbers from among $\{1,2,3,4,5,6,7,8\}$.
Then I get to pick two of the five numbers you picked. If I pick two that add up to $9$, then I win; if the two I pick don't add up to $9$, then you win.
What the problem says is that I can always win, no matter which five numbers you pick during your turn.
In other words, you get to pick the five numbers, but you don't also get to specify which two you want to add up to 9.
The easiest example is to do $10$. Then break up your set into $5$ sets, $$\{1,\dots,20\}\\\{21,\dots 40\}\\\vdots\\\{81,\dots,100\}$$
Since you have selected $55$ elements, you must have selected $11$ elements from one of these partitions. Then from the hint, you are done.
The case for $10$ is easy because $20$ divides $100$. The case of $12$ is harder. Partition the numbers as:
$$\{1,\dots,24\}\\\{25,\dots,48\}\\\{49,\dots,72\}\\\{73,\dots, 96\}\\\{97,98,99,100\}$$
Now we know that we've selected at least $51=55-4$ from the first four sets, and hence at least $13$ from one of them, and again we are done by the hint.
The problem for $11$ is that the "remainder" is large. When we partition $100$ into sets of $22$, we are left with $12$ elements. That's a large enough number of elements left to "break" the logic above - we can pick up to $11$ different elements from that remainder, and we have $44$ to parcel out amongst the first four partitions, which we can do by putting $11$ in each partition. It is not hard to find a counter-example for $11$ precisely this way.
More generally, given $N$ and $k$, let $m=k\left\lfloor \frac{N}{2k}\right\rfloor$. Then you can select from $\{1,\dots,N\}$ the following number of elements:$$m + \min(k,N-2m)$$
and not get a pair that differ by $k$. But it you try to select more than that number, you must get a pair with difference $k$.
In the case $N=100$ and $k=11$, $m=44$ and $\min(k,N-2m)=11$ so we can select $55$ such numbers.
Best Answer
Alternative perspective:
Suppose that you have a set $S$ with exactly $(2k)$ elements, where $k$ is some positive integer. Further suppose that the set $S$ is partitioned into $k$ subsets, where each subset has $2$ elements. Here, the word partition signifies that:
Now suppose that you are required to select $(k+1)$ distinct elements from the set $S$. Is it the case that one of the $(k)$ subsets will have both of its elements selected?
Yes, because if you are constrained by only picking (at most) $1$ element from each of the $k$ subsets, then it will only be possible to select $k$ different elements. That is, when you then select the $(k+1)$-th element, it will have to come from one of the subsets from which you already selected an element.
This is (loosely) referred to as the pigeonhole principle. That is, if you have $(k+1)$ pigeons and $k$ pigeonholes, then $2$ of the pigeons have to be placed in the same pigeonhole.
You asked what is special about the number $(4)$ as it pertains to the posed problem. The problem specifically requires that $5 = (4+1)$ elements from $\{1,2,\cdots,8\}$ be selected.
This implies that if $\{1,2,\cdots,8\}$ can be partitioned into $(4)$ subsets of $2$ elements each, in a meaningful way, then it will be relevant that you were forced to select both elements from one of the $(4)$ subsets.
This is specifically because you are being required to select $(4+1)$ elements.
Then, the partitioning into
$\{1,8\}$
$\{2,7\}$
$\{3,6\}$
$\{4,5\}$
is specifically crafted so that when you select both elements from one of the $4$ subsets above, you will necessarily have selected some pair of elements that add up to $9$.
However, this all begs the question:
how was the problem originally crafted?
Answer:
It was recognized that the set $\{1,2,\cdots,8\}$ permitted the partitioning into the $4$ subsets shown above, such that in any of the $4$ subsets, the pair of numbers added up to $9$.
Then, once this was recognized, the problem composer realized that if you were forced to select $(4+1)$ different elements, then as discussed, you would be forced to select both elements from one of the $4$ subsets.