The simple answer to your question is that your choice of "pigeons" and "holes" didn't support a proof of the claim. That's not to say it didn't do something. As the commenters correctly pointed out, you showed that you can't have a set of $50$ numbers less than $99$ such that any two numbers in the $50$ sum to a unique number, relative to any other pair. If this is what the question had been asking, then your logic is absolutely sound.
But the quesiton wasn't asking that. It was asking for the existence of $99$ as one of the pairwise sums. And, in this case, the choice of pigeons and holes given in the book works, and yours doesn't.
The thing about the pigeon hole principle, or any other proof technique, is that it's a tool. And like any tool, it has to be applied the right way to solve the problem at hand. This is a great lesson for you as a mathematician- to see that there are many ways to apply a single tool to a single problem. It's not enough to say "The pigeon hole principle didn't work". You must say "The pigeon hole principle, with this choice of pigeons and holes, gives me this result, which doesn't seem to help." All the same, you mentally note it in case you need it later, and then you either change your choice of tool or try to apply the tool in a different way. Problem solving is an art, as much as a science.
First off, your method of choosing some numbers is incorrect. $50$ and $51$ do add to $101$, but if $k=2$, then the question would say that whatever two numbers we pick, they would sum to $101$. But if we say, chose $1$ and $2$, these clearly don't add to $101$. What we want is no matter which $k$ numbers we pick, we can find two that sum to $101$.
The best way to think about this is to think not of individual numbers, but of pairs.
What I mean by this, is instead of $100$ numbers from $1$ to $100$, we now have $50$ pairs
$$\{(1,100), (2,99), (3,98), ..., (50,51) \}$$
And now, it should become apparent that in order to choose no pairs that sum to $101$, we must choose a number from each pair at most once. Thus, when we choose numbers from $51$ pairs, by pidgeonhole principle, we will have chosen both numbers from some pair, and thus have two numbers adding to $101$. Hope that helps!
Best Answer
The primes below 10 are 4:
$2,3,5,7$
So each of your numbers has the form:
$2^a \cdot 3^b \cdot 5^c \cdot 7^d$
There 16 possible tuples (a,b,c,d) when each of $a,b,c,d$ is viewed modulo 2.
We define each pigeonhole as the sequence $(A,B,C,D)$ where A,B,C,D are the residues of a,b,c,d respectively when divided by 2. There are 16 possible pigeonholes ($2 \cdot 2 \cdot 2 \cdot 2$).
This means you can find two numbers $X,Y$ among those 17 such that their tuples
$a_1,b_1,c_1,d_1$ and $a_2,b_2,c_2,d_2$
are such that
$a_1$ and $a_2$ are both odd or both even
$b_1$ and $b_2$ are both odd or both even
$c_1$ and $c_2$ are both odd or both even
$d_1$ and $d_2$ are both odd or both even
Now multiply X and Y and you get a square because
$a_1 + a_2$, $b_1 + b_2$, $c_1 + c_2$, $d_1 + d_2$
will all be even.