After posting a bad solution yesterday I believe I got a better one:
The strategy of the buyer consists of two functions, $(f_1(v,p_1),f_2(v,p_1,p_2))$ where both functions map to $\left\{A,R\right\}$ (where $A$ stands for Accept, $R$ for Reject).
The strategy of the seller is $(p_1,p_2(f_1(v,p_1)))$. You get the solution via backward induction. In PBE $f_2(v,p_1,p_2)$ maps to $A$ if and only if $v \geq p_2$. (There is inconsequential leeway at equality.) In PBE the seller believes that there is a set $H$ of types for which the buyer refused her offer $p_1$. Then
$$
p_2^* = \arg\max_{p_2} p_2 \cdot Prob(f_2(v,p_1,p_2) = A | f_1(v,p_1) = R).
$$
The buyer will accept offer $p_1$ if and only if
$$
v - p_1 \geq \delta \cdot (v - p_2).
$$
From this you get
$$
v \cdot (1 - \delta) \geq p_1 - \delta \cdot p_2.
$$
The left hand side of this equation is increasing in $v$, so types with high valuation will Accept. This means that in PBE the set $H$ is such that
$$
H = [0, \bar{v}).
$$
From this we get the optimal $p_2$ given $\bar{v}$:
$$
p_2^* = \arg\max_{p_2} p_2 \cdot Prob(v \geq p_2 | v \in [0, \bar{v})) = \frac{\bar{v}}{2}.
$$
In PBE $\bar{v}$ is a function of $p_1$:
$$
\bar{v} \cdot (1 - \delta) = p_1 - \delta \cdot \frac{\bar{v}}{2},
$$
so
$$
\bar{v} = \frac{p_1}{1 - \frac{\delta}{2}}.
$$
We have determined all the PBE strategies but $p_1$.
The expected payoff of the seller is
$$
p_1 \cdot \left( 1 - \frac{p_1 - \delta \cdot p_2(\bar{v}(p_1))}{1 - \delta} \right) + \frac{1}{2} \cdot p_2(\bar{v}(p_1)) \cdot \left( \frac{p_1 - \delta \cdot p_2(\bar{v}(p_1))}{1 - \delta} - p_2(\bar{v}(p_1)) \right),
$$
where
$$
p_2(\bar{v}(p_1)) = \frac{\bar{v}(p_1)}{2} = \frac{\frac{p_1}{1 - \frac{\delta}{2}}}{2} = \frac{p_1}{2 - \delta}.
$$
Substituting this we get
$$
p_1 \cdot \left( 1 - \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} \right) + \frac{1}{2} \cdot \frac{p_1}{2 - \delta} \cdot \left( \frac{p_1 - \delta \cdot \frac{p_1}{2 - \delta}}{1 - \delta} - \frac{p_1}{2 - \delta} \right),
$$
You have to maximize this w.r.t. $p_1$. With $\delta = 0.5$ I got
$$
p_1^* = \frac{9}{20}, \hskip 20pt \bar{v} = \frac{3}{5}, \hskip 20pt p_2^* = \frac{3}{10}.
$$