$W(n)$ is the function that counts number of distinct prime divisors of $n$. I have been able to prove for any $m$ consecutive integers starting with $1+a$ with the condition $a\leq (m^2-4m)/4$ , there exist a number $n$ in that sequence with the property $W(n)\leq 2$.
Is it worth to publishing? Is it some thing new?
On
After investigating a big list of sequence A001221, I found $$\omega(30684)=\omega(30685)=\omega(30686)=\omega(30687)=\omega(30688)=3,$$
and
$$\omega(n)=3 \text{ for } 99843 \leq n \leq 99850,$$
hence Option 3 of the other answer turns out to be false and there is little evidence that increasing $m$ - say $m>8$ - might really help us.
Short answer: no, don't publish this. If you want to publish anything, you should first make sure you've stated the theorem properly.
As has been discussed in the comments, the theorem was a little unclear. But you've explained what theorem you actually meant, so let's state it once more to avoid any confusion.
This is trivial: I can give you any sequence starting at a prime, for example, $$23,24,\cdots,23+m-1$$ and that is such a sequence (since the first number of the sequence, in this case, $23$, has $\omega(23)=1$).
However, let's state the other two options here.
or
The third option is disproved by MooS and Patrick Stevens by counterexamples (see MooS's answer or Patrick Stevens' comment).
Option 2 is also disproved by Patrick Stevens, by cleverly noting that any sequence of $30$ consecutive integers contains at least one multiple of $30$, and so at least one number in that sequence has at least $3$ prime factors.