The job is to use Pigeonhole Principle and prove that you can choose $12$ different 2-digit integers whose difference will be a 2-digit number with the same ciphers $(11, 22, 33, ..., 99)$.
I've reached a point where I figured that there are 90 possible differences: The smallest one being $99-98$ or $11-10$ $= 1$ and the largest being $99-10=89$ which sums up to $90$ possible differences.
And I'm stuck here
Let \begin{align*}a&=11k+\alpha,\\ b&=11l+\beta,\end{align*} where $k,l,\alpha,\beta\in\mathbb N,\,\alpha,\beta<11,\, a>b$.Then, $a-b = 11(k-l)+(\alpha-\beta)$, therefore if $a-b$ is divisible by $11$, $\alpha-\beta=0\implies \alpha=\beta$. So for the difference of $\alpha$ and $\beta$ to be a multiple of eleven, $\alpha$ and $\beta$ must have the same remainder when divided by eleven.
We can write every integer between 10 and 99 in the form $11k + \alpha$, where $k,\alpha\in\mathbb N,\,0\le k \le 9,\,0\le \alpha\le 10$. So we can pick 11 numbers, which will all have a unique remainder when divided by $11$, however we must pick $12$. Therefore, at least two numbers will share the same remainder when divided by $11$ and as a result, their difference will be a multiple of $11$.