It appears that \begin{equation*} S_k=\frac23+\frac ak+\frac b{k^{3/2}}+\frac c{k^2}+\frac{d_k}{k^{5/2}},\tag{1}\label{1} \end{equation*} where $a,b,c$ are certain real numbers such that $a>0$ and the $|d_k|$'s are bounded by a certain real $d$.

A proof of \eqref{1} should be rather straightforward (even if quite tedious) using Stirling's formula and the [Laplace method][1] -- both with higher-order terms but with explicit bounds on the remainder terms everywhere, as well as the Euler--Maclaurin summation formula. The first steps in this direction can be the observations that (i) \begin{equation*} S_k=\sum_{i=1}^k a_{k,i} \end{equation*} with
\begin{equation*} a_{k,i}:=\frac{i (2 k-i-1)!}{(2 i-k)! (k-i)! (2 k-2 i)!}\Big/ \binom{2 k-1}{k} \end{equation*} and (ii) $a_{k,i+1}\ge a_{k,i}$ if $i\le\frac23\,k-1$ and $a_{k,i+1}\le a_{k,i}$ if $i\ge\frac23\,k$.

It will then follow from \eqref{1} that $S_k$ is decreasing in $k\ge k_{a,b,c,d}$, where $k_{a,b,c,d}$ depends only on $a,b,c,d$. If $k_{a,b,c,d}$ is not too large, it should then be easy to check that $S_k$ is decreasing in $k$ if $3\le k\le k_{a,b,c,d}$. Thus, you will get that $S_k$ is decreasing in all $k\ge3$, to $2/3$, as desired.

It appears that \begin{equation*} S_k=\frac23+\frac ak+\frac b{k^{3/2}}+\frac c{k^2}+\frac{d_k}{k^{5/2}},\tag{1}\label{1} \end{equation*} where $a,b,c$ are certain real numbers such that $a>0$ and the $|d_k|$'s are bounded by a certain real $d$.

A proof of \eqref{1} should be rather straightforward (even if quite tedious) using Stirling's formula and the Laplace method -- both with higher-order terms but with explicit bounds on the remainder terms everywhere, as well as the Euler--Maclaurin summation formula. The first steps in this direction can be the observations that (i) \begin{equation*} S_k=\sum_{i=1}^k a_{k,i} \end{equation*} with
\begin{equation*} a_{k,i}:=\frac{i (2 k-i-1)!}{(2 i-k)! (k-i)! (2 k-2 i)!}\Big/ \binom{2 k-1}{k} \end{equation*} and (ii) $a_{k,i+1}\ge a_{k,i}$ if $i\le\frac23\,k-1$ and $a_{k,i+1}\le a_{k,i}$ if $i\ge\frac23\,k$.

It will then follow from \eqref{1} that $S_k$ is decreasing in $k\ge k_{a,b,c,d}$, where $k_{a,b,c,d}$ depends only on $a,b,c,d$. If $k_{a,b,c,d}$ is not too large, it should then be easy to check that $S_k$ is decreasing in $k$ if $3\le k\le k_{a,b,c,d}$. Thus, you will get that $S_k$ is decreasing in all $k\ge3$, to $2/3$, as desired.

It appears that \begin{equation*} S_k=\frac23+\frac ak+\frac b{k^{3/2}}+\frac c{k^2}+\frac{d_k}{k^{5/2}},\tag{1}\label{1} \end{equation*} where $a,b,c$ are certain real numbers such that $a>0$ and the $|d_k|$'s are bounded by a certain real $d$.

A proof of \eqref{1} should be rather straightforward (even if quite tedious) using Stirling's formula and the Laplace method -- both with higher-order terms but with explicit remainder terms everywhere, as well as the Euler--Maclaurin summation formula. The first steps in this direction can be the observations that (i) \begin{equation*} S_k=\sum_{i=1}^k a_{k,i} \end{equation*} with
\begin{equation*} a_{k,i}:=\frac{i (2 k-i-1)!}{(2 i-k)! (k-i)! (2 k-2 i)!}\Big/ \binom{2 k-1}{k} \end{equation*} and (ii) $a_{k,i+1}\ge a_{k,i}$ if $i\le\frac23\,k-1$ and $a_{k,i+1}\le a_{k,i}$ if $i\ge\frac23\,k$.

It will then follow from \eqref{1} that $S_k$ is decreasing in $k\ge k_{a,b,c,d}$, where $k_{a,b,c,d}$ depends only on $a,b,c,d$. If $k_{a,b,c,d}$ is not too large, it should then be easy to check that $S_k$ is decreasing in $k$ if $3\le k\le k_{a,b,c,d}$.

It appears that \begin{equation*} S_k=\frac23+\frac ak+\frac b{k^{3/2}}+\frac c{k^2}+\frac{d_k}{k^{5/2}},\tag{1}\label{1} \end{equation*} where $a,b,c$ are certain real numbers such that $a>0$ and the $|d_k|$'s are bounded by a certain real $d$.

A proof of \eqref{1} should be rather straightforward (even if quite tedious) using Stirling's formula and the [Laplace method][1] -- both with higher-order terms but with explicit bounds on the remainder terms everywhere, as well as the Euler--Maclaurin summation formula. The first steps in this direction can be the observations that (i) \begin{equation*} S_k=\sum_{i=1}^k a_{k,i} \end{equation*} with
\begin{equation*} a_{k,i}:=\frac{i (2 k-i-1)!}{(2 i-k)! (k-i)! (2 k-2 i)!}\Big/ \binom{2 k-1}{k} \end{equation*} and (ii) $a_{k,i+1}\ge a_{k,i}$ if $i\le\frac23\,k-1$ and $a_{k,i+1}\le a_{k,i}$ if $i\ge\frac23\,k$.

It will then follow from \eqref{1} that $S_k$ is decreasing in $k\ge k_{a,b,c,d}$, where $k_{a,b,c,d}$ depends only on $a,b,c,d$. If $k_{a,b,c,d}$ is not too large, it should then be easy to check that $S_k$ is decreasing in $k$ if $3\le k\le k_{a,b,c,d}$. Thus, you will get that $S_k$ is decreasing in all $k\ge3$, to $2/3$, as desired.