You've made a good start. The only problem is that $\binom{2n}n$ only counts the number of paths from $(0,0)$ to $(n,n)$. You can combine each of these with any path from $(n,n)$ to $(2n,2n)$, and there are also $\binom{2n}n$ of those, so the total number of paths going through $(n,n)$ is $\binom{2n}n^2$.
If you're not sure about something like this, a good idea is to try it out concretely for small numbers. In the present case, you can very easily enumerate all the paths for $n=1$ and find that $4$ of the $6$ go through $(1,1)$. If you'd done this, you'd probably have noticed what was missing in your argument.
There are a few ways to solve this question, each of which uses a different, but very useful, counting principle. We wish to count the paths that stay weakly below the line $y=x$ (they can touch the line, but not go above)
1) Reflection Principle
First, we consider the total number of paths that exist from $(0,0)$ to $(n,n)$, which is ${2n}\choose{n}$, since we must take exactly $n$ up steps, each of which we denote by a $u$, and $n$ right steps, each of which we denote by an $r$, and so this simply corresponds to the ${2n}\choose{n}$ ways we can rearrange a string with $n$ $u$s and $n$ $r$s.
Next, we consider the paths that DO go above the line $y=x$ (the bad paths). We note that for each such path, there exists a unique first upstep that brings the path above the line (it could even be the first step). We draw a line of slope $1$ passing through the end of this unique up step, and reflect the path up to that step about the line, giving a path from $(-1,1)$ to $(n,n)$. (See picture)
Now, we've shown that any bad path can be mapped onto a path from $(-1,1)$ to $(n,n)$. It is indeed true (but you should convince yourself of this) that any path from $(-1,1)$ to $(n,n)$ can be mapped onto a bad path from from $(0,0)$ to $(n,n)$. Hence, there are the same number of bad paths from from $(0,0)$ to $(n,n)$ as the number of all paths from $(-1,1)$ to $(n,n)$, which by the argument above is ${2n}\choose{n-1}$=${2n}\choose{n+1}$.
Hence the number of paths that stay below the line $y=x$ is (after some manipulation): $${{2n}\choose{n}} - {{2n}\choose{n-1}}=\dfrac{1}{n+1}{{2n}\choose{n}}$$
2) Generating Functions
If you are comfortable with generating functions, we can let $g(x)$ be the generating function for paths that stay below the line $y=x$. Let us call paths that stay below the line good paths. Then we can note that each path that is below the line $y=x$ can be broken into two after the first return to the axis.
Now if we let $x^{1/2}$ be the weight of each step, then the first part of the path can be written as $x g(x)$, as it is a right step ($x^{1/2}$), followed by a good path ($g(x)$), followed be an up step ($x^{1/2}$). The second part of the path is just another good path ($g(x)$). The only path that cannot be written this way is the lone good path that does not start with a right step, namely, the empty path from $(0,0)$ to $(0,0)$, who's generating series is just $x^0=1$. Hence we have that
$$g(x) = x g(x) \cdot g(x)+1=x g(x)^2+1 $$
Solving for $g(x)$ gives:
$$g(x) = \dfrac{1 \pm \sqrt{1-4x} }{2x}$$
One can then use the generalized binomial theorem to find the series:
$$g(x) = \sum_{n=0}^{\infty} a_n x^n$$
and you should get $a_n = \dfrac{1}{n+1} {{2n}\choose{n}}$, and of course by our construction, $a_n$ counts the good paths from $(0,0)$ to $(n,n)$.
3) Cycle Lemma
Using the Cycle Lemma (http://www.cs.tau.ac.il/~nachumd/papers/CL.pdf)
one can map the good paths from $(0,0)$ to $(n,n)$ to dominating strings of $n+1$ $0$s and $n$ $1$s, where $0$s represent right steps and $1$s represent up steps. The cycle lemma tells us that of the $2n+1$ cyclic permutation of each such possible string, only $(n+1)-n=1$ of them are dominating. and hence since there are ${2n+1}\choose{n}$ possible strings, the total number of dominating strings, which corresponds to the good paths, is:
$$\dfrac{1}{2n+1} {{2n+1}\choose{n}} = \dfrac{1}{n+1} {{2n}\choose{n}}$$
Best Answer
Yeah, there is a nice way to do it. This looks long, but it's because I stated everything rigorously. If you draw pictures while reading this, it'll make a LOT more sense.
Let $f(2n)$ denote the number of paths from $(0, 0)$ to $(2n, 2n)$ not crossing through a point of the form $(2k+1, 2k+1)$. I claim that $f(2n) = C_{2n}$, where $C_{2n}$ is the $2n$-th Catalan number.
A well known property of Catalan number $C_{n}$ is that it satisfies the following recursion formula: $$ C_{n+1} = \sum_{i=0}^{n} C_i C_{n-i} \tag{1}$$ Another well known property is that it counts the number of paths from $(0,0)$ to $(2n,2n)$ which never go above the line $y=x$.
I'll prove the result by induction. Notice it's true for a base case of $n = 0$. Now, suppose the result is true for $f(0), f(2), \dots, f(2n-2)$.
To count $f(2n)$, we do casework on the first point of the form $(2k, 2k)$ our path goes through (other than $(0, 0)$). This casework covers all paths since all paths end up at $(2n, 2n)$. Suppose the first such point is $(2k, 2k)$. WLOG on our first step, we went $(0, 0) \to (1, 0)$, we'll multiply by $2$ in our final count. Then we must also end with $(2k, 2k-1) \to (2k, 2k)$. It remains to count the number of paths which go from $(1, 0)$ to $(2k, 2k-1)$ without passing any point of the form $(2k, 2k)$. This is just $C_{2k-1}$! After this, there are $f(2n-2k)$ ways to finish up the path $(2k, 2k) \to (2n, 2n)$. Therefore, we have $$f(2n) = \sum_{k=1}^{n} 2 \cdot C_{2k-1} f(2n-2k)$$ By the inductive hypothesis, $f(2n-2k) = C_{2n-2k}$, so we really have $$f(2n) = \sum_{k=1}^{n} 2 \cdot C_{2k-1} C_{2n-2k} = \sum_{k=1}^n C_{2k-1}C_{2n-2k} + \sum_{k=1}^nC_{2k-1}C_{2n-2k}$$ using $j = n-k$ as the iterator for the second sum, we get $$f(2n) = \sum_{k=1}^n C_{2k-1}C_{2n-2k} + \sum_{j = 0}^{n-1} C_{2j} C_{2n-2j}$$ The finish is in sight! The first sum is just $C_1C_{2n-2}+C_3C_{2n-4} + \dots C_{2n-1}C_{0}$ (i.e. the odd terms from $(1)$) while the second sum is just $C_{0}C_{2n-1} + \dots C_{2n-2}C_1$ (i.e. the even terms from $(1)$). Therefore, we deduce that $f(2n) = C_{2n}$ as desired.
I'm sure the bijective proof exists, but I have yet to try to find it. But given this, maybe you'll be able to do it :)