Let's prove it by contradiction, and suppose at some moment, the beetle can't return to the starting cell anymore. Let $S$ denote the set of cells that the beetle can go to at this moment. For any adjacent $a\in S, b\in S^c$, since the beetle can go to $a$ but not $b$, the door should be open and the direction the door is open is from $b$ to $a$. $\Rightarrow$ no door is open in the direction from $S$ to $S^c$. That is, the beetle goes from $S^c$ to $S$ for only one time any never leaves $S$. $\Rightarrow$ the beetle can only open one door between $S$ and $S^c$, but the number of doors between them $>1$. $\therefore$ some doors between $S$ and $S^c$ are closed, contradiction, and this finished the proof.

If at a moment beetle wants to go where she started then beetled should go to a square on the diagonal of the checkered square then she can use diagonal symmetry by doing the moves she had done before by reflecting them along the diagonal