For case 3, consider $B_0 \setminus U_1$ and $B_0\setminus U_2$. But you get it more directly, without any case distinction, by: Let $x_1 \neq x_2$. Then $\exists U_1,U_2$ ... Then consider $B_1 = A\cap U_1$. You will find that $B_1 \in f(x_1)\setminus f(x_2)$, so $f(x_1) \neq f(x_2)$.

@DanielFischer, thank you very much! I've got it now!

That's great. Did you do it both ways or only one?

I did it with the more direct method you suggested; it seemed more elegant.

@DanielFischer I typed up a solution below to illustrate how I used the direct method. If you are available to provide some more advice, there's a different topology problem I'm stuck on that I just posted here: math.stackexchange.com/q/757255/111520

Arthur Fischer (no relation) gave a good hint there meanwhile. If you need further hints, feel free to ping, but I think you will understand what he intends.

Hmm.. I'm aware that the intersection of finitely many open sets is open, and I'd sure like to construct a new nested countable neighborhood basis from the new one. But... I'm still not sure what direction he was trying to point me in.

here's an idea:

I'm not the only slow typist ;)

Given the countable neighborhood basis \{U_\alpha\}_{\alpha \in \mathbb N} Define a recursive sequence of open sets \{V_k\}_{k \in \mathbb N}, with the first set V_1 defined as U_1, and then each set V_{k+1} defined as the intersection of V_k and U_{k+1}???

How shall I put it? Bingo!

Haha, thank you! My only concern is that V_{k+1} might not be a proper subset of V_k, which I thought would be the ideal version of a "nested set". (I think my construction only guarantees subset or equal)

Yes, that only guarantees $\subseteq$, not $\subsetneq$, but $\subsetneq$ may not be possible, the point may have a smallest neighbourhood.

Okay, that's true. I was holding my expectations up to a standard beyond what is possible!

For example in a discrete space, $\{x\}$ must appear in a neighbourhood basis, if $V_k = \{x\}$, then $V_n = \{x\}$ for all $n \geqslant k$.

Okay, that makes perfect sense. The only way this problem would be soluble, then, is if my definition of nested allows $\subseteq$, and does not require $\subsetneq$.

The usual definition of nested allows $\subseteq$.

Okay; I didn't realize that. Well I have to go off to class, but thank you very much for your advice!

You're welcome.