A generating function proof.

As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$ we have that \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}&= [z^{2n}] \frac{\arcsin(z)}{z} \frac{1}{\sqrt{1-z^2}}\\ &= [z^{2n+1}]\arcsin(z)\,\arcsin^\prime(z)\\ &=(2n+2) [z^{2n+2}] \frac{1}{2} \big(\arcsin(z)\big)^2 \end{align*} The series expansion of $\frac{1}{2} \big(\arcsin(z)\big)^2$ was already given by Euler and is well known \begin{align*} \frac{1}{2} \big(\arcsin(z)\big)^2=\sum_{n\geq 0} \frac{4^n (n!)^2}{(2n+2)!}z^{2n+2}\end{align*} (See e.g. formula 1.645.1 in Gradshteyn-Ryzhik). Thus \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}=\frac{4^n}{(2n+1){2n \choose n}},\,\mbox{ as claimed.}\end{align*} (Of course, this may also be seen as a special case of hypergeometric series summation.)

ADDED: the Taylor expansion of $y(x)=\frac{1}{2}\big(\arcsin(x))^2$ can be derived independently from the conjectured equality, by using that $y$ solves the differential equation \begin{align*} (1-x^2)y^{\prime\prime} - xy^\prime=1\end{align*} with $y(0)=y^\prime(0)=0$, and undetermined coefficients. (This is in fact what Euler did).

A generating function proof.

As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$ we have that \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}&= [z^{2n}] \frac{\arcsin(z)}{z} \frac{1}{\sqrt{1-z^2}}\\ &= [z^{2n+1}]\arcsin(z)\,\arcsin^\prime(z)\\ &=(2n+2) [z^{2n+2}] \frac{1}{2} \big(\arcsin(z)\big)^2 \end{align*} The series expansion of $\frac{1}{2} \big(\arcsin(z)\big)^2$ was already given by Euler and is well known \begin{align*} \frac{1}{2} \big(\arcsin(z)\big)^2=\sum_{n\geq 0} \frac{4^n (n!)^2}{(2n+2)!}z^{2n+2}\end{align*} (See e.g. formula 1.645.1 in Gradshteyn-Ryzhik). Thus \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}=\frac{4^n}{(2n+1){2n \choose n}},\,\mbox{ as claimed.}\end{align*} (Of course, this may also be seen as a special case of hypergeometric series summation.)

ADDED: the Taylor expansion of $y(x)=\frac{1}{2}\big(\arcsin(x)\big)^2$ can be derived independently from the conjectured equality, by noting that $y$ solves the differential equation \begin{align*} (1-x^2)y^{\prime\prime} - xy^\prime=1\end{align*} with $y(0)=y^\prime(0)=0$, and using undetermined coefficients. (This is in fact what Euler did).

A generating function proof.

As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$ we have that \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}&= [z^{2n}] \frac{\arcsin(z)}{z} \frac{1}{\sqrt{1-z^2}}\\ &= [z^{2n+1}]\arcsin(z)\,\arcsin^\prime(z)\\ &=(2n+2) [z^{2n+2}] \frac{1}{2} \big(\arcsin(z)\big)^2 \end{align*} The series expansion of $\frac{1}{2} \big(\arcsin(z)\big)^2$ was already given by Euler and is well known \begin{align*} \frac{1}{2} \big(\arcsin(z)\big)^2=\sum_{n\geq 0} \frac{4^n (n!)^2}{(2n+2)!}z^{2n+2}\end{align*} (See e.g. formula 1.645.1 in Gradshteyn-Ryzhik). Thus \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}=\frac{4^n}{(2n+1){2n \choose n}},\,\mbox{ as claimed.}\end{align*}

 (Of course, this may also be seen as a special case of hypergeometric series summation.)

A generating function proof.

As $\arcsin(z)=\sum_{k\geq 0} \frac{1}{2k+1} {2k \choose k} \frac{z^{2k+1}}{4^k}$ and $\frac{1}{\sqrt{1-z^2}}=\sum_{k\geq 0}{2k \choose k}\frac{z^{2k}}{4^k}$ we have that \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}&= [z^{2n}] \frac{\arcsin(z)}{z} \frac{1}{\sqrt{1-z^2}}\\ &= [z^{2n+1}]\arcsin(z)\,\arcsin^\prime(z)\\ &=(2n+2) [z^{2n+2}] \frac{1}{2} \big(\arcsin(z)\big)^2 \end{align*} The series expansion of $\frac{1}{2} \big(\arcsin(z)\big)^2$ was already given by Euler and is well known \begin{align*} \frac{1}{2} \big(\arcsin(z)\big)^2=\sum_{n\geq 0} \frac{4^n (n!)^2}{(2n+2)!}z^{2n+2}\end{align*} (See e.g. formula 1.645.1 in Gradshteyn-Ryzhik). Thus \begin{align*} \frac{1}{4^n} \sum_{k=0}^n \frac{1}{2k+1}{ 2k \choose k}{2(n-k) \choose n-k}=\frac{4^n}{(2n+1){2n \choose n}},\,\mbox{ as claimed.}\end{align*}  (Of course, this may also be seen as a special case of hypergeometric series summation.)

ADDED: the Taylor expansion of $y(x)=\frac{1}{2}\big(\arcsin(x))^2$ can be derived independently from the conjectured equality, by using that $y$ solves the differential equation \begin{align*} (1-x^2)y^{\prime\prime} - xy^\prime=1\end{align*} with $y(0)=y^\prime(0)=0$, and undetermined coefficients. (This is in fact what Euler did).