All odd primes, except for $3$, are of the form $6n \pm 1$.
So, if two odd primes bigger than $3$ exists and differ by $2$, they must be of the form
$6n-1, 6n+1$ for some integer $n$.
The number that are $2$ before and $2$ after will show up as
$6n-3, 6n-1, 6n+1, 6n+3$
If $n=1$, you get $3,5,7,9$ and the first three numbers are prime.
For any $n>1$, only the middle two numbers might be prime.
A simpler proof.
Let $(x, x+2, x+4)$ be three consecutive odd numbers.
Modulo 3, those three numbers will have to look like one of
(0, 2, 1) or (1, 0, 2) or (2, 1, 0)
Hence at least one of them will be a multiple of three.
Since the number 3 is a prime number, we need to look at those cases where one of those three numbers is 3.
(-1, 1, 3) or
(1, 3, 5) or
(3, 5, 7)
So (3, 5, 7) is the only sequence of three consecutive odd prime numbers.