All other prime gaps (except between $2$ and $3$) are even.

$1$. The only even prime is $2$ and all other are odd primes.

$2$. Difference between two odd numbers is always even.

First posted as hint but now @Cato asked for a proof so here We go:

Proof of $1$:

Suppose that there is another prime which is even (other than $2$), then it must be of the form $2k$ where $k$ being a positive integer, so that prime number can be written as $2 \times k$ which is enough to say that it is composite (Or not prime). A contradiction.

Proof of $2$.

As known (I don't think it also need a proof), every odd number is of the form $2k+1$, where $k$ is an integer. We just have to calculate the difference b/w two odd primes (Remember the word odd) so let first one be $2k_1+1$and second one be $2k_2+1$. Difference $= (2k_2+1-(2k_1+1))= 2k_2-2k_1=2(k_2-k_1)=$ even number.

I think it is enough. :) :)

Since all prime numbers except $2$ are odd, then if a prime gap $g$ between primes $p_1,p_2$ would be odd, then $p_2=p_1+g$ would be the sum of two odd numbers which is even, and since the only prime which is even equals $2$, thus we get the desired contradiction.

All primes other than $2$ are odd, hence gap between them is even (as difference of two odd numbers is always even).