Proving an identity relating to binomial coefficients This question is from the book Discrete mathematics and its applications, by K. Rosen, 6th chapter and 27th question.

Show that if n is a positive
integers $2C(2n, n+1) + 2C(2n, n) = 
  C(2n+2,n+1)$

I know how to prove this algebraically, applying the formula and this is how it is also given in hints. Since I want to understand counting arguments I am thinking to give a proof of this using combinatorial arguments (double counting).
The right side is the number of ways to select n+1 elements from 2n+2 elements. But I am not able to get an argument for the left side. I'm not getting how can I argue that I need to choose n or n+1 elements from 2n elements twice.
Any other (counting) approaches are also fine. Thank you.

Here we show the validity of the binomial identity
\begin{align*}
\color{blue}{2\binom{2n}{n+1}+2\binom{2n}{n}=\binom{2n+2}{n+1}}\tag{1}
\end{align*}
by counting lattice paths. We consider lattice paths with horizontal steps $(1,0)$ and vertical steps $(0,1)$. The binomial coefficient
\begin{align*}
\binom{n}{k}\tag{2}
\end{align*}
gives the number of lattice paths of length $n$ from $(0,0)$ to $(k,n-k)$ consisting of $k$ horizontal $(1,0)$ steps and $n-k$ vertical $(0,1)$ steps.
Let's have a look at the graphic below. We see an $(n+1)\times(n+1)$ rectangle where we study lattice paths from the origin $(0,0)$ to $(n+1,n+1)$.
                                                
We observe

*

*The number of lattice paths from $(0,0)$ to $(n+1,n+1)$ is according to (2) equal to
\begin{align*}
\color{blue}{\binom{2n+2}{n+1}}\tag{3.1}
\end{align*}


*A lattice path from $(0,0)$ to $(n+1,n+1)$ has to cross either the blue point $(n,n)$ or one of the orange points $(n-1,n+1)$ or $(n+1,n-1)$ respectively. This way we can partition the lattice paths into three sets.

*

*Blue point $(n,n)$: The number of lattice path from $(0,0)$ to the blue point $(n,n)$ is $\binom{2n}{n}$.  From the blue point $(n,n)$ there are two ways to go to $(n+1,n+1)$. Either by a horizontal step $(n,n)\to(n+1,n)\to (n+1,n+1)$ or by a vertical step $(n,n)\to(n,n+1)\to(n+1,n+1)$. So we have two possibilities giving a total of
\begin{align*}
\color{blue}{2\binom{2n}{n}}\tag{3.2}
\end{align*}
lattice paths.


*Orange point $(n+1,n-1)$: The number of lattice paths from $(0,0)$ to the orange point $(n+1,n-1)$ is
\begin{align*}
\binom{2n}{n-1}=\color{blue}{\binom{2n}{n+1}}\tag{3.3}
\end{align*}
From $(n+1,n-1)$ there is just one possibility to go to $(n+1,n+1)$: namely $(n+1,n-1)\to(n+1,n)\to(n+1,n+1)$. The total number of lattice paths from $(0,0)$ to $(n+1,n+1)$ crossing $(n+1,n-1)$ is therefore $\binom{2n}{n-1}=\binom{2n}{n+1}$.


*Orange point $(n-1,n+1)$: A symmetrical situation. The number of lattice paths from $(0,0)$ to the orange point $(n-1,n+1)$ is
\begin{align*}
\color{blue}{\binom{2n}{n+1}}\tag{3.4}
\end{align*}
Since there is just one possibility to go from $(n-1,n+1)$ to $(n+1,n+1)$: namely $(n-1,n+1)\to(n,n+1)\to(n+1,n+1)$, the total number of lattice paths from $(0,0)$ to $(n+1,n+1)$ crossing $(n-1,n+1)$ is $\binom{2n}{n+1}$.

Putting (3.1) to (3.4) together we see
\begin{align*}
\color{blue}{2\binom{2n}{n+1}+2\binom{2n}{n}=\binom{2n+2}{n+1}}
\end{align*}
and the claim (1) follows.

