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