Open neighbourhoods, Polish spaces, and basis for the Baire space This is a follow-up of a question I asked yesterday answered by GEdgar. I think I see now GEdgar’s answer, but I am not sure about an issue related to it. Thus, I will write my general understanding of the topic, something I developed by thinking – hopefully in the right way – about GEdgar’s answers (plural, because he kindly gave me a feedback to my comments|questions written below is formal answer). Then I will ask my new question.
[I write my understanding in a rather pedantic way, but useful for me – and hopefully for you as readers – to see the logic behind each step clearly. This is also due to the fact that topology is really difficult for me to properly grasp, and the product topology is one of the most problematic things for me to see.]
1. Let $\mathbb{N}$, and  endow it with the discrete topology. Then build $\mathcal{N} := \mathbb{N}^\mathbb{N} \equiv \mathbb{N}^\omega$. By looking at the natural topology of $\mathcal{N}$, i.e. the product topology on it, we get the open sets of $\mathcal{N}$.
2. Let $\sigma \in \mathbb{N}^{<\omega}$.
3.  Here, due to the product topology, every $\{ \sigma \} \subset \mathbb{N}^{<\omega}$ is open (this is relevant to get the logic behind point (5)).
4. By fixing $\sigma$ we can build $N_\sigma := \{ \ f \in \mathcal{N} \ | \ \sigma \subset f \ \}$.
5. Now, $N_\sigma$ is an open neighbourhood of $f$ because it is made of all the other $f$ that share $\sigma$, and every $\{ f \}$ is itself open by the same logic applied to $\sigma$ at point (3), being thus also a subset of $N_\sigma$.
6. We focus on $N_\sigma$ simply because this is an example of a set in the basis for the topology of $\mathcal{N}$. 
QUESTIONS:
 - How does an open neighbourhood of $\sigma \in \mathbb{N}^{<\omega}$ look like?
 - Does the logic behind point (6) applies to this open neighbourhood as well? 
My guess is that first we fix $\mathbb{N}^{<\omega}$ to be a definite finite number $n$, by getting $\mathbb{N}^n$. Then we let $\sigma^* \in \mathbb{N}^m$, with $m < n$. Finally, we apply the same line of reasoning  at point (4) and (5).
Am I right (overall)?
Any feedback regarding my general understanding, or this specific question, is greatly welcome.
Thank you for your time.
 A: I’ll address both the numbered points and the issues that came up in the comments; for convenience I repeat the former here.

1. Let $\mathbb{N}$, and  endow it with the discrete topology. Then build $\mathcal{N} := \mathbb{N}^\mathbb{N} \equiv \mathbb{N}^\omega$. By looking at the natural topology of $\mathcal{N}$, i.e. the product topology on it, we get the open sets of $\mathcal{N}$.
2. Let $\sigma \in \mathbb{N}^{<\omega}$.
3.  Here, due to the product topology, every $\{ \sigma \} \subset \mathbb{N}^{<\omega}$ is open (this is relevant to get the logic behind point (5)).
4. By fixing $\sigma$ we can build $N_\sigma := \{ \ f \in \mathcal{N} \ | \ \sigma \subset f \ \}$.
5. Now, $N_\sigma$ is an open neighbourhood of $f$ because it is made of all the other $f$ that share $\sigma$, and every $\{ f \}$ is itself open by the same logic applied to $\sigma$ at point (3), being thus also a subset of $N_\sigma$.
6. We focus on $N_\sigma$ simply because this is an example of a set in the basis for the topology of $\mathcal{N}$. 

(1) and (2) are fine, but (3) makes no sense: $\Bbb N^{<\omega}$ isn’t a product and therefore does not have a product topology. We do not in fact endow it with any topology; this answers your first question.
As Asaf noted in the comments, we can embed $\Bbb N^{<\omega}$ into the product $\Bbb N^\omega$ in a natural way by appending an infinite sequence of zeroes to each $\sigma\in\Bbb N^{<\omega}$. If, for instance, $\sigma=\langle m_0,\ldots,m_{n-1}\rangle$, we can define $\hat\sigma=\langle m_k:k\in\omega\rangle\in\Bbb N^\omega$, where $m_k=0$ for all $k\ge n$. 
Let $D=\{\hat\sigma:\sigma\in\Bbb N^{<\omega}\}$; then as Asaf said, $D$ is a countable subset of $\Bbb N^\omega$ that has no isolated points. $D$ is not countable because it’s the product of finitely many copies of $\Bbb N$: $D$ is not a product at all, and neither is $\Bbb N^{<\omega}$. $D$ is countable because there is a bijection between $D$ and $\Bbb N^{<\omega}$, and the latter is countable: $\Bbb N^{<\omega}=\bigcup_{n\in\omega}\Bbb N^n$, and each $\Bbb N^n$, being the product of finitely many copies of $\Bbb N$, is countable.
To see that $D$ has no isolated points, let $\hat\sigma\in D$ be arbitrary. Then $\left\{N_{\hat\sigma\upharpoonright n}:n\in\omega\right\}$ is a local base of open sets at $\hat\sigma$ in $\Bbb N^\omega$ (where I’m using the notation of your (4)). For $n\in\omega$ let $\tau\in\Bbb N^{n+1}$ be defined by setting $\tau\upharpoonright n=\hat\sigma\upharpoonright n$ and $\tau(n)=\hat\sigma(n)+1$; then $\hat\tau\in D\cap N_{\hat\sigma\upharpoonright n}$, but $\hat\tau\ne\hat\sigma$.
Finally, $D$ is metrizable, being a subspace of the metrizable space $\Bbb N^\omega$, so $D$ is homeomorphic to $\Bbb Q$. (That $D$ is totally disconnected follows from this, but we don’t need total disconnectedness to get this.)
I suspect that Asaf brought up this subspace $D$ of $\Bbb N^\omega$ because it’s the most natural space related to the non-space $\Bbb N^{<\omega}$ of your (3), and it’s very far from being a discrete space: no singleton is open in $D$.
Going on to (5), $N_\sigma$ is an open nbhd of each of its elements, but the argument that you give is simply wrong: no $\{f\}$ is an open set. If $\sigma=\langle m_0,\ldots,m_{n-1}\rangle$, say, let $U_k=\{m_k\}$ for $k<n$; $\Bbb N$ has the discrete topology, so each $U_k$ is open. Then
$$N_\sigma=\left\{f\in\Bbb N^\omega:f(k)\in U_k\text{ for each }k<n\right\}$$
is a basic open ‘box’ in the product $\Bbb N^\omega$.
The sets of the form $N_\sigma$ aren’t just a base for the topology of $\Bbb N^\omega$: they turn out to be an exceptionally useful base, one that is especially easy to work with. (It is sometimes very handy, for instance, that the partial order $\left\langle\{N_\sigma:\sigma\in\Bbb N^{<\omega}\},\supseteq\right\rangle$ is a tree.)
