Infinitely repeated game with stationary and symmetric equilibrium
Why don't you attack the problem straighrforwardly and investigate the profitability of a deviation?
Suppose there was a Nash equilibrium in which both players decide to stay forever. Both players obtain payoff $\frac{1}{1-d}$ on path.
This really is a Nash equilibrium if anf only if there is no profitable deviation. Consider the deviation to quit in the first period, yielding the deviation payoff $10$. If a later deviation to quitting or a smaller probability to quit was profitable then so would be this pure earliest deviation.
Hence, always staying is a Nash equilibrium when $\frac{1}{1-d}\geq 10 \iff d\geq 0.9$.