Alternative simple variant:
$$
 \int_0^{1-a}\frac{H(z)}{1-z}dz=\int_0^{1-a}\frac1{(1-z)^2}\int_z^1\frac{2\zeta}{1+\zeta}H(\zeta^2)d\zeta=
$$
$$
 \frac1{1-z}\int_z^1\frac{2\zeta}{1+\zeta}H(\zeta^2)d\zeta|_{z=0}^{z=1-a}+\int_0^{1-a}\frac{2\zeta}{1-\zeta^2}H(\zeta^2)d\zeta,
$$
which leads to
$$
 \int_{1-2a-a^2}^{1-a}\frac{H(z)}{1-z}dz\to H(1)-H(0)\ \ \ for\ \ \ a\to0.
$$
Since
$$
 \int_{1-2a}^{1-a}\frac{dz}{1-z}\to\ln2,
$$
we obtain the result.