Let $(\Bbb{X}, \rho)$ be a Metric Space. Consider the Metric Topology $(\Bbb{X}, \tau_{\rho})$ and fix $x \in \Bbb{X}$. Why does $\mathscr{B}_x = \{B_{\rho}(x, \epsilon): \epsilon > 0\}$ form, in general, an uncountable Local Bases at $x$, but $\mathscr{B}_x = \{B_{\rho}(x, \frac{1}{n}): \frac{1}{n} < \epsilon \text{ where } n \in \mathbb{N}^+\}$ forms a countable Local Bases at $x$?
I understand why both $\mathscr{B}_x$ form Local Bases at $x$, and I understand that $\mathbb{N}^+$ is a countable set, hence, the radii $\frac{1}{n}$ are countable. However, I am having trouble really understanding what is going on here.
EDIT: I think I may have thought this through...Moreover, consider the Standard Topology. Fix $x \in \Bbb{R}$, then the two Local Bases are the following:
$$\mathscr{B}_x = \{(x - \frac{1}{n}, x + \frac{1}{n}): \frac{1}{n} < \epsilon \text{ where } n \in \mathbb{N}^+\} \tag{1}$$
$$\mathscr{B}_x = \{(x - \epsilon, x + \epsilon): \epsilon > 0\} \tag{2}$$
So what (2) is saying is that for some fixed $x \in \Bbb{R}$, say $x = 0$, then for every open set containing $x = 0$, say $(-1,1)$, we can fit the open interval $(x - \epsilon, x + \epsilon): \epsilon > 0$, i.e. $(- \epsilon,+ \epsilon): \epsilon > 0$. Hence, there are uncountable many $\epsilon > 0$ that do so...just for that one open set about $x = 0$. For (1) it is the same logic, however, you have countable many sets $(- \frac{1}{n},+ \frac{1}{n})$...Is this the correct way of thinking?
You have two bases at $x$, let us denote them by two different symbol to distinguish them:$\newcommand{\ve}{\varepsilon}\newcommand{\R}{\mathbb R}\newcommand{\N}{\mathbb R}\newcommand{\abs}[1]{|#1|}$ \begin{gather*} \mathscr B_x=\{B(x,\ve); \ve>0, \ve\in\R\}\\ \mathscr B'_x=\{B(x,\frac1n); n\in\N^+\} \end{gather*} Notice that I have omitted $\frac1n<\ve$ which you have in the question. Each set belonging to $\mathscr B'_x$ is determined by $n$, it does not depend on an additional parameter $\ve$.
In the case of real line with the usual metric, we will get \begin{gather*} \mathscr B_x=\{(x-\ve,x+\ve; \ve>0, \ve\in\R)\}\\ \mathscr B'_x=\{x-\frac1n, x+\frac1n); n\in\N^+\} \end{gather*}
Why they are countable/uncountable? Let us consider, for simplicity $x=0$. (If we look at a different point, then the whole situation is just shift.)
If we look at $\mathscr B_x$, we can see that for each $\ve\in(0,\infty)$ we get a different set. (If $0<\ve_1<\ve_2$ then the point $\frac{\ve_1+\ve_2}2$ belongs to $(-\ve_2,\ve_2)$ but does not belong to $(-\ve_1,\ve_1)$.) Therefore $\abs{\mathscr B_x}\ge \abs{(0,\infty)}$ cardinality of $\mathscr B_x$ is at least the cardinality of $(0,\infty)$. Therefore this set is uncountable.
On the other hand, for $\mathscr B'_x$ we have a bijection between $\N^+$ and $\mathscr B'_x$. Thus $\mathscr B'_x$ is countable.
Why is it a basis? Again, we will look at $x=0$.
We know that for every open set $U$ containing zero there is small enough $\ve>0$ such that $(-\ve,\ve)\subseteq U$. (This is the fact that $\mathscr B_x$ is a neighborhood basis at $x$.)
For every $\ve>0$ there exists large enough $n\in\N^+$ such that $\frac1n<\ve$. This means that we have an element from $\mathscr B'_x$ such that $$0\in (-\frac1n,\frac1n) \subseteq (-\ve,\ve) \subseteq U.$$ And we see that every open neighborhood of zero contains also some set from $\mathscr B'_x$.