A nice but somewhat challenging binomial identity 
When working on a problem I was faced with the following binomial identity valid for integers $m,n\geq 0$:
  \begin{align*}
\color{blue}{\sum_{l=0}^m(-4)^l\binom{m}{l}\binom{2l}{l}^{-1}
\sum_{k=0}^n\frac{(-4)^k}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\binom{k+l}{l}
=\frac{1}{2n+1-2m}}\tag{1}
\end{align*}
I have troubles to prove it and so I'm kindly asking for support.

Maybe  the following  simpler one-dimensional identity could be useful for a proof. We have for non-negative integers $n$:
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}=\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}\tag{2}
\end{align*}
The  LHS of (2) can be transformed to
\begin{align*}
\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}\int_{0}^1x^{2k}dx\\
&=\int_{0}^1\sum_{k=0}^n(-1)^k\binom{n}{k}x^{2k}\,dx\\
&=\int_{0}^1(1-x^2)^n\,dx
\end{align*}
Using a well-known integral representation of reciprocals of binomial coefficients the RHS of (2) can be written as
\begin{align*}
\frac{4^{n}}{2n+1}\binom{2n}{n}^{-1}&=4^n\int_{0}^1x^n(1-x)^n\,dx
\end{align*}
and the equality of both integrals can be shown easily. From (2) we can derive a simple one-dimensional variant of (1).
We consider binomial inverse pairs and with respect to (2) we obtain
\begin{align*}
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}g_k \quad&\quad g_n=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&f_n=\sum_{k=0}^n(-1)^k\binom{n}{k}\frac{1}{2k+1} \quad&\quad\frac{1}{2n+1}=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k
\end{align*}

We conclude again with (2)
  \begin{align*}
\frac{1}{2n+1}&=\sum_{k=0}^n(-1)^k\binom{n}{k}f_k\\
&=\sum_{k=0}^n\frac{(-4)^{k}}{2k+1}\binom{n}{k}\binom{2k}{k}^{-1}\\
\end{align*}
  This identity looks somewhat like a one-dimensional version of (1). Maybe this information can be used to solve (1).

 A: Let us complete the OP's work, started with
$$ \frac{1}{2k+1}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k}{(2k+1)\binom{2k}{k}}\tag{$d=0$}$$
by computing first the binomial transform of $\frac{1}{2k+3}$. We have:
$$\begin{eqnarray*}\sum_{k=0}^{n}\frac{(-1)^k}{2k+3}\binom{n}{k}=\int_{0}^{1}x^2(1-x^2)^n=\frac{B\left(n+1,\tfrac{3}{2}\right)}{2}=\frac{1}{2n+3}\cdot\frac{B\left(n+1,\frac{1}{2}\right)}{2}\end{eqnarray*}$$
hence:
$$ \frac{1}{2k+3}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k}{(2k+1)(2k+3)\binom{2k}{k}}\tag{$d=1$}$$
and in general:
$$ \frac{1}{2k+2d+1}\stackrel{\text{Binomial transform}}{\longleftrightarrow} \frac{4^k\binom{k+d}{d}\binom{2k}{k}^{-1}}{(2k+2d+1)\binom{2k+2d}{2d}}\tag{$d\geq 1$}$$
I need some time to check the above computations, but the last identity, together with creative telescoping, should be the key for proving OP's statement. Indeed, we have:
$$ \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}=\frac{1}{2n+1}\tag{$l=0$} $$
$$ \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}(k+1)=-\frac{1}{(2n+1)(2n-1)}\tag{$l=1$} $$
$$\begin{eqnarray*} \sum_{k=0}^{n}\frac{(-4)^k}{(2k+1)\binom{2k}{k}}\binom{n}{k}\binom{k+l}{l}&=&\frac{(-1)^l(2l-1)!!(2n-2l+1)!! }{(2n+1)!!}\\
&=&\frac{(-1)^l 4^{n-l} n! (2l)! (n-l)!}{(2n+1)!l! (2n-2l+1)!}\tag{$l\geq 1$} \end{eqnarray*}$$
hence the whole problem boils down to computing:
$$ \frac{4^n}{(2n+1)\binom{2n}{n}}\sum_{l=0}^{m}\frac{\binom{m}{l}}{(2n-2l+1)!\binom{n}{l}}$$
A: This is by no means an answer, but may help. Equation (6.28) here is most likely a corollary of Vandermonde's identity with appropriate values for the parameters, but it's too late for me to figure out what they are. This reduces your sum to
$$\frac{2^{2n}}{(2n+1)}\binom{2n}{n}^{-1}\sum_{l=0}^m (-4)^l \binom{m}{l}\binom{2l}{l}^{-1}\binom{n-l-\frac{1}{2}}{n}.$$
By the way, Mathematica can evaluate this sum, giving (almost) your right-hand side.
A: Firstly let us evaluate the inner sum on the left hand side. Using the beta function identity quoted above along with the identity $\left. \binom{k+l}{l} = d^l/dx^l x^{k+l}/l! \right|_{x=1}$ we have:
\begin{equation}
S^{(n)}_l:=\sum\limits_{k=0}^n \frac{(-4)^k}{2k+1} \binom{n}{k} [\binom{2k}{k}]^{-1} \binom{k+l}{l} = \left.\frac{1}{l!} \frac{d^l}{d x^l} x^l \int\limits_0^1 \left(1- 4 t (1-t) x\right)^n dt \right|_{x=1}
\end{equation}
Now if we take $m=0$ then $l=0$ and then:
\begin{equation}
rhs= 4^n \int\limits_0^1 \left[ (t-\frac{1}{2})^2 \right]^n dt= 4^n \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} u^{2 n} du = \frac{1}{2 n+1}
\end{equation}
as it should be.
Now let us take arbitrary $l \ge 0$ . Then by using the chain rule of differentiation and then by substituting $u := t-1/2$ we have:
\begin{equation}
S^{(n)}_l= \sum\limits_{p=0}^l \binom{l}{p} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p}(4 u^2-1)^p d u
\end{equation}
Therefore the left hand side of the identity to be proved reads:
\begin{eqnarray}
&&\sum\limits_{l=0}^m (-4)^l \binom{m}{l} [\binom{2 l}{l}]^{-1} S^{(n)}_l=\\
&&\sum\limits_{p=0}^m(-1)^{p+1} 2^{2p-1} \frac{\binom{m}{p} (m-p-3/2)!(p-1/2)!}{\sqrt{\pi} \binom{2 p}{p} (m-1/2)!} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p} (4 u^2-1)^p du=\\
&&\sum\limits_{p=0}^m (-1)^{p+1} 4^p \frac{\binom{m}{p} \binom{m}{1/2}}{\binom{2 p}{p} \binom{m}{p+3/2}} \cdot \frac{1}{(2p+1)(2p+3)} \binom{n}{p} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (4 u^2)^{n-p} (4 u^2-1)^p du=\\
&&-4^n \sum\limits_{p=0}^m \binom{m}{p} \frac{\binom{n}{p} \binom{m}{1/2}}{\binom{2 p}{p} \binom{m}{p+3/2}} \cdot \frac{1}{(2p+1)(2p+3)} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} u^{2n-2p} (1-4 u^2)^p du=\\
&&-4^n \frac{1}{2} \frac{n!}{(m-1/2)!} \sum\limits_{p=0}^m  \binom{m}{p} \frac{(m-p-3/2)!}{(n-p)!} \int\limits_{-\frac{1}{2}}^{\frac{1}{2}} (u^2)^{n-p} (1/4 - u^2)^p d u=\\
&& -\frac{1}{4} \frac{n! m!}{(n+1/2)! (m-1/2)!} \sum\limits_{p=0}^m \frac{(m-p-3/2)!(n-p-1/2)!}{(n-p)!(m-p)!}=\\
&& -\frac{1}{(2m-1)(2n+1)} F^{3,2}\left[\begin{array}{rrr} 1&-m&-n\\ \frac{3}{2}-m & \frac{1}{2}-n & \end{array};1\right] = \\
&& -\frac{1}{(2m-1)(2n+1)} \cdot \frac{(\frac{1}{2}-m)^{(n)} (\frac{3}{2})^{(n)}}{(\frac{3}{2}-m)^{(n)} (\frac{1}{2})^{(n)}} = \\
&&-\frac{1}{(2m-1)(2n+1)} \cdot  \frac{(1- 2m)(1+2 n)}{1-2 m+2 n} = \frac{1}{2n-2 m+1}
\end{eqnarray}
where in the first line we summed over $l$ and in the second in the third and in the forth  lines we simplified the result. Finally in the fifth line we evaluated the integral by substituting for $4 u^2$ and in the subsequent line we expressed the sum through hypergeometric functions. Finally, from Wolfram's site http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric3F2/03/02/04/,  we used the following identity:
\begin{equation}
F^{(3,2)}\left[ \begin{array}{rrr} a& b & -n \\ d & a+b-d-n+1 & \end{array};1 \right] = \frac{(d-a)^{(n)}(d-b)^{(n)}}{(d)^{(n)} (-a-b+d)^{(n)}}
\end{equation}
for $a=1$, $b=-m$ and $d=3/2-m$.
