We consider lattice paths starting from $(0,0)$ containing horizontal $(1,0)$-steps and vertical $(0,1)$-steps only, with $x<y$ at all other points besides the origin.
The following is valid. The number of paths of length $2n$ starting at $(0,0)$ and never touching the diagonal $x=y$ again is
\begin{align*}
\binom{2n-1}{n-1}\qquad\qquad\qquad n\geq 1
\end{align*}
Note, the number of paths of shortest length from $(x_0,y_0)$ to $(x_1,y_1)$ with $x_0\leq x_1$ and $y_0\leq y_1$ is
\begin{align*}
\binom{(x_1-x_0)+(y_1-y_0)}{x_1-x_0}\tag{1}
\end{align*}
Paths of length $2n$ starting at $(0,0)$ and never touching the diagonal again end in points
\begin{align*}
(k,2n-k)\qquad\qquad\qquad 0\leq k \leq n-1
\end{align*}
Since the first step is always from $(0,0)$ to $(0,1)$ in order to walk above the diagonal, we consider at first the number of all paths from $(0,1)$ to $(k,2n-k)$. This number is according to (1)
\begin{align*}
\binom{2n-1}{k}\qquad\qquad\qquad 0\leq k \leq n-1
\end{align*}
From this number we have to subtract the number of all paths from $(0,1)$ to $(k,2n-k)$ which touch (or cross) the diagonal $x=y$.
This can be determined using Andre's Reflection Principle:
A path starting from $(0,1)$ going to $(k,2n-k)$ which touches the diagonal, will touch it the first time let's say at $(t,t)$. So, each such path consists of two parts. The first part from $(0,1)$ to $(t,t)$ and the second from $(t,t)$ to $(k,2n-k)$. We bijectively map paths of this shape by reflecting the first part at the diagonal and leaving the second part unchanged. This way we get a reflected path starting at $(1,0)$ crossing at $(t,t)$ the diagonal the first time and ending in $(k,2n-k)$.
We conclude: The number of paths from $(0,1)$ to $(k,2n-k)$ which touch the diagonal $x=y$ is equal to the number of all pathes from $(1,0)$ to $(k,2n-k)$ and is according to (1)
\begin{align*}
\binom{2n-1}{k-1}
\end{align*}
Let's denote with $P_k(n)$ the number of all paths from $(0,0)$ to $(k,2n-k)$ which do not touch the diagonal again. The following is valid for $n\geq 1$
\begin{align*}
P_k(n)=
\begin{cases}
1\qquad\qquad &k=0\\
\binom{2n-1}{k}-\binom{2n-1}{k-1}\qquad\qquad &k\geq 1
\end{cases}\tag{2}
\end{align*}
Note that for $k=0$ there is exactly one path from $(0,0)$ to $(0,2n)$. Since in this case there are no paths touching the diagonal which are to subtract, we interpret the binomial coeffient $\binom{r}{s}$ with negative $s$ as usual with zero and we obtain $P_0(n)=\binom{2n-1}{0}-\binom{2n-1}{-1}=1$.
Now it's time to harvest: The number of all paths of length $2n$ starting at $(0,0)$ and never touching the diagonal again is
\begin{align*}
\sum_{k=0}^{n-1}P_k(n)&=\sum_{k=0}^{n-1}\left(\binom{2n-1}{k}-\binom{2n-1}{k-1}\right)\\
&=\sum_{k=0}^{n-1}\binom{2n-1}{k}-\sum_{k=1}^{n-1}\binom{2n-1}{k-1}\tag{3}\\
&=\sum_{k=0}^{n-1}\binom{2n-1}{k}-\sum_{k=0}^{n-2}\binom{2n-1}{k}\\
&=\binom{2n-1}{n-1}
\end{align*}
and the claim is proved.
Comment: In (3) we split the sum and skip the index $k=0$ in the right hand sum since it's contribution is zero. In the following line we shift the index by $1$ in the right hand sum.
