Q: Continuity of a function? 
Let $f: (\mathbb N, e_\mathbb N) \to(\mathbb R,e)$, where $e$ stands for the euclidean metric and
  $$f(n)=\begin{cases} n,\, n\ge 2\\\\0,\, n=1\end{cases}$$
    Is $f$ continuous?

Firstly, I can prove that $d_0,\,  e_\mathbb N$ are topologically equivalent in $\mathbb N$, which means that every $A\subset \mathbb N$ is $d_0-$ open and $e_\mathbb N-$ open.
Secondly, we consider an open set $B \in (\mathbb R,e)$. 
If $f$ is continuous, then $f^{-1}(B)$ must be open in  ($\mathbb N, e_\mathbb N$).
However,  $f^{-1}(B)= \{ x \in \mathbb N: f(x) \in B\}\subset \mathbb N$, which means $f^{-1}(B)$ is open in $(\mathbb N, e_\mathbb N)$. Thus, we have that $f$ is continuous.
Is it right?
Edited

A: Hint: A function is continuous if the inverse image of any open set is open.
Consider which subsets of $\mathbb{N}$ are open.
