A hausdorff topology that is not regular Consider, for $a$ and $b$ positive integers, the set $U(a,b)=\{an+b\mid n\in \mathbb{Z}\}\cap \mathbb{N}$. The set $\{U(a,b)\mid (a;b)=1\}$ is a basis for some topology on the naturals (I proved this). 

Show that this space is Hausdorff but not regular. 

Take $n, m\in \mathbb{N}$ such that $m\neq n$. I don't know hot to find disjoint open sets for each. Any hints?

To prove it's not regular, I tried taking the point $1$ and the set $C=\{2n\mid n\in \mathbb{N}\}$. Any open set $W$ such that $1\in W$ will have a basic set contained in it, but I don't see the problem.  

Any help would be really appreciated.
For Hausdorff: 

Consider $U(mn+1, m), U(mn+1,n)$. Since $\gcd(a\cdot k + 1, a) =1$, these are both basis elements, hence open. Further, they're disjoint, as if there exists $j,k$ such that
$$ (mn+1)k+m = (mn+1)j+n$$
Then $m = n $ modulo $mn+1$, but $m, n$ are distinct, both positive and less than $mn+1$, so this cannot happen. Thus there are no such $j,k$, and the sets are disjoint.

Now, for regular. Take $C,p$ as you've outlined above. Suppose for contradiction there are disjoint open sets $A, B$ containing $p, C$ respectively. Note that $A$ must contain a set of the form $U(k_1,1)$, where $k_1$ is even (otherwise $k_1+1 \in C$ and $k_1+1 \in U(k_1,1)$), as the only other basis sets containing $1$ also contain all of $\mathbb{N}$ - which obviously intersects with $B$.

Likewise, $B$ must contain contain $U(k_2,k_1)$ (with $gcd(k_2,k_1) = 1$) as it must contain $k_1$ (a positive even number, which is in $C$). Again if we have some other open set $U(a,b)$ containing $k_1$, then $a*n + b = k_1$ for some $n$, and hence this open set contains $U(a, a*n+b) = U(k_2, k_1)$.

However, $U(k_1,1) \cap U(k_2,k_1) \neq \emptyset$, because of Bezouts identity, since $gcd(k_2,k_1) = 1$, we have $i,j \in \mathbb{Z}$ such that $$i k_2 + j k_1 = 1 \implies i k_2 + k_1 = 1 - (j-1) k_1$$

To turn this element into a natural number (so we get our intersection) add a sufficiently large multiple of $k_1 \cdot k_2$ to each side. This is clearly an element of both $U(k_2, k_1)$ and $U(k_1, 1)$, so $A,B$ are disjoint. A contradiction.
For Hausdorff: 

Consider $U(mn+1, m), U(mn+1,n)$. Since $\gcd(a\cdot k + 1, a) =1$, these are both basis elements, hence open. Further, they're disjoint, as if there exists $j,k$ such that
$$ (mn+1)k+m = (mn+1)j+n$$
Then $m = n $ modulo $mn+1$, but $m, n$ are distinct, both positive and less than $mn+1$, so this cannot happen. Thus there are no such $j,k$, and the sets are disjoint.

Now, for regular. Take $C,p$ as you've outlined above. Suppose for contradiction there are disjoint open sets $A, B$ containing $p, C$ respectively. Note that $A$ must contain a set of the form $U(k_1,1)$, where $k_1$ is even (otherwise $k_1+1 \in C$ and $k_1+1 \in U(k_1,1)$), as the only other basis sets containing $1$ also contain all of $\mathbb{N}$ - which obviously intersects with $B$.

Likewise, $B$ must contain contain $U(k_2,k_1)$ (with $gcd(k_2,k_1) = 1$) as it must contain $k_1$ (a positive even number, which is in $C$). Again if we have some other open set $U(a,b)$ containing $k_1$, then $a*n + b = k_1$ for some $n$, and hence this open set contains $U(a, a*n+b) = U(k_2, k_1)$.

However, $U(k_1,1) \cap U(k_2,k_1) \neq \emptyset$, because of Bezouts identity, since $gcd(k_2,k_1) = 1$, we have $i,j \in \mathbb{Z}$ such that $$i k_2 + j k_1 = 1 \implies i k_2 + k_1 = 1 - (j-1) k_1$$

To turn this element into a natural number (so we get our intersection) add a sufficiently large multiple of $k_1 \cdot k_2$ to each side. This is clearly an element of both $U(k_2, k_1)$ and $U(k_1, 1)$, so $A,B$ are disjoint. A contradiction.