Decompose your lattice paths in two parts: the one up until reaching $(3,3)$ and from $(3,3)$ to $(10, 10)$ not passing by $(5,5)$. We can translate this second part to be actually the number of paths from $(0, 0)$ to $(7,7)$ not passing by $(2,2)$. The fact that your paths must pass through $(3,3)$ make these problems independent.
For the first one we have ${6 \choose 3}$ possibilities
For the second one, the number of paths not passing through $(5,5)$ is all the paths minus those that do pass through $(5,5)$. We can calculate the second number using the same strategy splitting the path from $(3,3)$ to $(5,5)$ and then from $(5,5)$ to $(10, 10)$. This can be done in $${4 \choose 2}{10 \choose 5}$$ ways. So in total we get, for this second part $${14 \choose 7} - {4 \choose 2}{10 \choose 5}$$.
Since we can choose any path for the first part and combine it with any other one from the second, the nuumber we get is the product of the number for those parts which is $${6 \choose 3}\left({14 \choose 7} - {4 \choose 2}{10 \choose 5}\right)$$
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.
Best Answer
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.