Prove that palindromes are not regular language without using pumping lemma
You can use the Myhill–Nerode criterion.

Suppose that $A$ is a DFA accepting your language. Let $q_n$ be the state which the DFA reaches upon reading the word $0^n1$. As the DFA has finitely many states, we must have $q_n = q_m$ for some $n \neq m$. On the one hand, when at state $q_n$, upon reading $0^n$ the DFA must reach an accepting state, since $0^n 1 0^n$ is a palindrome. On the other hand, when at state $q_m$, upon reading $0^n$ the DFA must reach a non-accepting state, since $0^m 1 0^n$ is not a palindrome. We have reached a contradiction since $q_n = q_m$.