Generalizing $\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2$ I was looking at this paper on section [17],

$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)}=2\tag1$$

Let generalize $(1)$
$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)\cdots [2n-(2k+1)]}\tag2$$
Where $k\ge 0$
I conjectured the closed form for $(2)$ to be  

$$\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)\cdots [2n-(2k+1)]}=\frac{2(-1)^k}{(2k+1)!!(2k+1)}\tag3$$

Here are a first few values of $k=1,2$ and $3$
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)}&=-\frac{2}{9}\tag4\\
\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)}&=\frac{2}{75}\tag5\\
\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{2^{2n}(2n-1)(2n-3)(2n-5)(2n-7)}&=-\frac{2}{735}\tag6
\end{align}$$
How do we go about to prove this conjecture $(3)?$

Using Lemma $(13)$ in this paper provided by the OP
$$\sum_{n=1}^{\infty}H_n{2n \choose n}x^n=\frac{2}{\sqrt{1-4x}}\ln\left(\frac{1+\sqrt{1-4x}}{2\sqrt{1-4x}}\right)$$
Relpace $x$ with $\displaystyle \frac{x^2}{4}$, we get 
$$\sum_{n=1}^{\infty}\frac{H_n}{4^n}{2n \choose n}x^{2n}=\frac{2}{\sqrt{1-x^2}}\ln\left(\frac{1+\sqrt{1-x^2}}{2\sqrt{1-x^2}}\right)$$
Now divide both sides by $x^2$ and integrate from $x=0$ to $x=1$, we get
\begin{align}
S&=\sum_{n=1}^{\infty}\frac{H_n{2n \choose n}}{4^n(2n-1)}=\int_0^1\frac{2}{x^2\sqrt{1-x^2}}\ln\left(\frac{1+\sqrt{1-x^2}}{2\sqrt{1-x^2}}\right)\ dx,\quad \color{blue}{x=\sin\theta}\\
&=2\int_0^{\pi/2}\csc^2\theta\ln\left(\frac{1+\cos\theta}{2\cos\theta}\right)\ d\theta\overset{IBP}{=}2\int_0^{\pi/2}\frac{d\theta}{1+\cos\theta}=2(1)=2
\end{align}
