There's no injective, continuous function $(\mathbb{Q},\lvert\cdot\rvert)\rightarrow(\mathbb{N},\lvert\cdot\rvert)$

Assume a continuous and injective $$g:(\mathbb{Q},\lvert\cdot\rvert)\rightarrow(\mathbb{N},\lvert\cdot\rvert)$$Then $$\forall A\subset\mathbb{N}\implies\overline{f^{-1}(A)}\subset\ f^{-1}(A)$$ (because every subset of $\mathbb{N}$ is closed and so $\overline{A}=A$).
That can only hold if:
$$\overline{f^{-1}(A)}=f^{-1}(A)$$
$\iff\ f^{-1}(A)$ is closed over $\mathbb{Q}$
I don't know how to continue. I haven't used the injective trait so that has to come into play but I'm stuck. I'm thinking the way to go would be finding a sequence in the preimage of A that converges to a point outside of it.