We number the doors by $1,2,3,...,n$. Our method is, basically, check the following sequence of doors in each iteration and we want to prove this guarantees to find the cat no matter how the cat move.
We prove the correctness of the method by using function property. We first show that if the cat starts in an even numbered door, we are guaranteed to find it during the first half of the traversal.
We let the number of iterations passed be our $x$ axis and the position of us be $f(x)$, the position of the cat be $g(x)$. The position is defined as the door number.
For example, using our traversal method, $$f(1)=2, f(2)=3, f(3)=4,..,f(n-2)=n-1$$
And $g(x)$ can be anything satisfying the following condition: $|g(x+1)-g(x)|=1$ for all $x$
Here comes the trick: note that they are discrete functions but we can approximate them with continuous functions. We extend $f(x)$ so that $f(x)=x+1$ for all real numbers $x$. We also extend $g(x)$ so that we keep the integer points and connect adjacent points with straight lines.
Notice the fact that $|g(x+1)-g(x)|=1$ for any integer $x$.
Now, since the cat starts in an even position, the following must be true.
(1)It cannot start at position $1$ and,
(2)After the $(n-2)$th iteration it cannot end at position $n$, because after $n-2$ iterations the cat has moved $n-3$ times (the first iteration the cat does not move). If $n$ is even $n-1$ is odd and the cat must end at odd position and if $n$ is odd similarly.
Now, we can conclude that $g(1)>=2$ and $g(n-2) <= n-1$
because we have extended $f(x)$ and $g(x)$ to continuous functions on real numbers in interval $[1,n-2]$, we can define $h(x)=f(x)-g(x)$ be another continuous function.
Note that $$h(1) = f(1) - g(1) >= 2 - 2 = 0$$ and $$h(n-2) = f(n-2)-g(n-2) <= n-1 - (n-1) = 0$$ so by intermediate value theorem, $h(x) = 0$ for some $c\in[1,n-2]$
We want to show that $c$ must be an integer. Suppose it is not can integer, then we have $a<c<a+1$ where $a$ and $a+1$ are consecutive integers. By definition of $h(c)=0$ we have $f(c)=g(c)$.
Note that $|g(a+1)-g(a)|=1$ and $f(a+1)-f(a)=1$. In order for the two function $f$ and $g$ to intersect at $x=c$, we must have $f(a)=g(a+1)$ and $g(a)=f(a+1)$. However, because $f(a)$ and $g(a)$ must be both even or both odd, this can't be true, so $c$ must be an integer.
Now we have proved the case if the cat starts at an even position then we mmust find it during our first traversal.
The second case is that if the cat starts at an odd position. In this case, we assume we haven't find the cat yet after the first traversal. Now we re-number the doors in reverse order and now the cat must be in an even position. It becomes an identical scenario to the first case.