Every 2 digit palindrome is obviously divisible by 11.
Suppose every 2n digit palindrome is divisible by 11.
Consider any 2(n+1) digit palindrome P. The central 2n digits define a 2n
digit palindrome PP divisible by 11, so write:
P=PP*10 + (10^(2n+1)+1)Q
where Q is the most (and least) significant digit of P. Then P is divisible by
11 because PP is and so is 10^(2n+1)+1.
Then the proof follows by mathematical induction.
(You may feel the need to prove that 10^(2n+1)+1 is divisible by 11.
This would proceed along the lines of observing that 10=-1 mod 11
so any odd power of 10 is also congruent to -1 modulo 11, which is
equivalent to any odd power of 10 plus 1 is divisible by 11)