$0 < 1 < \cdot\cdot\cdot < 9 < 11 < \cdot\cdot\cdot < 99 < \cdot\cdot\cdot n$. From this the possible values of $y$ are $1, 2, 11$. If $x_{i}$ has $n$ digits($n \geq 3$), then $x_{i+1}$ ends with the same digit as $x_{i}$, i.e it's divisible by 10, or not. In this case the two consecutive capicua can be of the form $x_{i} = a9...9a$, $x_{i+1} = b0...0b$ with $b = a + 1$ and $y_{i} = 11$. If $x_{i+1}$ has one more digit than $x_{i}$, they are of the form: $x_{i} = 9a...a9$, $x_{i+1} = 1b...b1$ with $y_{i} = 2$. Hence the set contains two distinct primes.

Given a capicua number, its easy to see that the most frequent way to find the next one is by changing the middle slot or slots by 1, for example if you had $12321$ the next capicua number is $12421$, if you had $257752$ then the next one would be $258852$ so on and so forth. Now, if it is of the form $99....99$ then the next one will be $100...01$ or if the middle spot cant go any further such is the case for $48799784$ for example, then you would have to change the slots closest to the middle spots and start over, in this particular case $48800884$, when you do that, the difference of those capicua numbers is just $11\cdot 10^2$. This process would repeat again for the 2nd pair of slots you changed, and they could eventually hit $9$ in which case if $x=$number of slots "untouched" to the right, you would have to add $11\cdot 10^x$ or $1\cdot 10^x$ if there is only 1 middle slot (i.e. if the capicua has an odd number of digits),this is just because we are repeating the case of the middle slots but we have to consider the "unused" decimal places of the number. Hence we have the only primes are $2$, $11$, since if any difference ends in $0$ it cant be a prime.