Let $\mathcal{H}_n$ denote the $n$-th harmonic number and consider the power series
$$\sum_{n=1}^{\infty} \mathcal{H}_n \mathcal{H}_{n+1} x^n \quad , \quad -1 \leq x <1$$
Since $\mathcal{H}_{n+1} = \mathcal{H}_n + \frac{1}{n+1}$ then we have that
\begin{align*}
\sum_{n=1}^{\infty} \mathcal{H}_n \mathcal{H}_{n+1} x^n &= \sum_{n=1}^{\infty} \mathcal{H}_n \left ( \mathcal{H}_n + \frac{1}{n+1} \right ) x^n \\
&=\sum_{n=1}^{\infty} \mathcal{H}_n^2 x^n + \sum_{n=1}^{\infty} \frac{\mathcal{H}_n}{n+1}x^n \\
&=\frac{\log^2 (1-x) +{\rm Li}_2(x)}{1-x} + \frac{\log^2(1-x)}{2x}
\end{align*}
Thus mapping $x \mapsto -x$ we get that
$$\sum_{n=1}^{\infty} (-1)^{n+1} \mathcal{H}_n \mathcal{H}_{n+1} x^n = - \frac{\log^2 (1+x)+{\rm Li}_2(-x)}{1+x} + \frac{\log^2(1+x)}{2x}$$
Integrating we get that
\begin{align*}
\int \sum_{n=1}^{\infty} (-1)^{n+1} \mathcal{H}_n \mathcal{H}_{n+1} x^n \, {\rm d}x&= \sum_{n=1}^{\infty} (-1)^{n+1} \mathcal{H}_n \mathcal{H}_{n+1} \int x^n \, {\rm d}x \\
&= \sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n \mathcal{H}_{n+1} x^{n+1}}{n+1}\\
&=\int \left ( \frac{\log^2 (1+x)+{\rm Li}_2(-x)}{1+x} + \frac{\log^2(1+x)}{2x} \right ) \, {\rm d}x\\
&= -3 {\rm Li}_3 (1+x) + {\rm Li}_2(-x) \log(1+x)+ 3{\rm Li}_2 (1+x) \log(1+x) \\
&\quad \quad + \frac{\log^3(1+x)}{3} + \frac{3}{2} \log(-x) \log^2 \left ( 1+x \right )
\end{align*}
Hence
$$\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{H}_n \mathcal{H}_{n+1} x^{n-1}}{n+1} = \frac{1}{x} \bigg[-3 {\rm Li}_3 (1+x) + {\rm Li}_2(-x) \log(1+x)+$$
$$+3{\rm Li}_2 (1+x) \log(1+x)+ \frac{\log^3(1+x)}{3} + \frac{3}{2} \log(-x) \log^2 \left ( 1+x \right ) \bigg] $$
Integrating from $0$ to $1$ we must get the result ....
There must be something more sufficient and clever here , no?