Proving that no two state DFA accept the language L = {$0^i : \forall i \ge 0$} $\lor$ {$^i : \forall i \ge 0$}
Use Myhill&ndash;Nerode theory. Here are four words which are pairwise incomparable:
$$ \epsilon,0,1,01. $$

The following table gives words separating them:
$$
\begin{array}{c|c|c|c|c}
&amp; \epsilon &amp; 0 &amp; 1 &amp; 01 \\\hline
\epsilon &amp; &amp; 1 &amp; 0 &amp; \epsilon\\\hline
0 &amp; 1 &amp; &amp; 0 &amp; \epsilon\\\hline
1 &amp; 0 &amp; 0 &amp; &amp; \epsilon\\\hline
01 &amp; \epsilon &amp; \epsilon &amp; \epsilon &amp; 
\end{array}
$$

According to Myhill&ndash;Nerode theory, every DFA for $L$ must contain at least four states. It is not too hard to construct a DFA containing four states, so four is the minimal number of states.