The answer is yes, almost trivially.
Prove the following lemma first:
Suppose that $B$ is a basis for a topology $\tau$. Define for $U\in\tau$ the set $B_U=\{V\in B\mid V\subseteq U\}$, then $U=U'$ if and only if $B_U=B_{U'}$.
Therefore you immediately get an injection from $\tau$ into $\mathcal P(B)$. And therefore if $|B|=|\Bbb N|$, you get that $|\tau|\leq|\mathcal P(\Bbb N)|$, as you wanted. Because you wanted at most, rather than "exactly".
The key point is to remember that there is often a lot of redundancy when you take a basis for a topology, and this is a feature rather than a bug. You want basis elements to become smaller and smaller, rather than having $U\mapsto B_U$ a bijection.
Here is a nice example, by the way, for "less". Take $\Bbb N$ with the topology that an open set is an initial segment of the order $\leq$. There are exactly $\aleph_0$ of those, and you need all of them (except $\varnothing$ and $\Bbb N$ themselves) to create a basis, too.
Or, you know, any topological space with finitely many open sets.
It's an infinite process, we have $B_1=B\setminus f(A)$, then
$$A_1=g(B_1),\ B_2=f(A_1),\ A_2=g(B_2),\ \dots, B_5=f(A_4),\ A_5=g(B_5),\ \dots$$
Yes, $(fg)^{i-1}$ is the $i-1$-fold composition of the composite $fg$ with itself. If $i-1=0$, we mean the identity function.
Best Answer
If you define the relation $\le$ on arbitrary sets by $$ |A|\le |B| :\iff \text{ there is an injection }A \to B $$ then you need the Cantor-Schröder-Bernstein theorem to prove (that is something you have to prove) that $\le$ is an order relation. Just because you call something $\le$, it does not mean that $$ |A|\le |B|, \ |B|\le |A| \implies |A|= |B| $$ holds true. That is Cantor-Schröder-Bernstein.