Problem concerning the use of Pigeonhole (Box) principle

Nice job, I think your reasoning is all there but it will help to break down your argument a little bit, to make it easier to read.

First you partitioned the interval $\{1,\cdots 112\}$ into $37$ sets, all of the form $S_x = \{x, x+9, x+19\}$ except for the exceptional set $X = \{19, 47, 75, 103\}$. Then we examine how $X$ interacts with the $S_x$. For any $z \in \{1,\cdots 112\}$, define $$B_z = \big\{y \in \{1,\cdots 112\} \text{ such that } |y-z| \in \{9,10,19\} \big\}$$

We make the important observation that for each $y \in X$, $B_y \supset S_{y-10}$. That is to say, for each element we place in $X$, we invalidate an additional $S_x$.

Now we have to place $37$ elements among the $37$ sets.

Suppose we place $0 \leq n \leq 4$ elements in $X$. This leaves $37 - n$ elements to place among the $36$ $S_x$. From the above observation, $n$ choices of the $S_x$ will immediately result in elements a 'bad' distance apart. So only $36 - n$ choices of the $S_x$ are viable. We are thus placing $37 - n$ elements in $36 - n$ of the $S_x$, and by pidgeonhole, one of the $S_x$ must contain two elements.

Just to show another possible way to tackle the problem.

Taking $S=\{1,\,2,\, \cdots,\,112\}$ or $S=\{0,\,1,\, \cdots,\,111\}$ or any set of $112$ consecutive integers clearly does not change the problem. Let's fix $S=\{0,\,1,\, \cdots,\,111\}$.

Then taking the set $T$ of $37$ integers, it is also clear that if we shift it as to have the minimum at $0$ does not change the problem.

Therefore we can start $T$ as $\{0,\, 1,\, 2,\, \cdots,\, 8\}$ (tot. of $9$ integers)
after which we shall jump the set $\{9,\, \cdots,\,18\} $ and $\{19,\, \cdots,\,27\} $ , i.e. $19$ integers.

The most compact way that we can proceed with is to repeat the above starting with $28$.

To reach to $|T|=37$ we need to fix $4$ banned sets, i.e. $76$ integers.
But $112-76=36 < 37$, so we cannot avoid that at least one integer be in banned set.