Subset of $\mathbb{R}$ without a countable neighborhood base An exercise in Ronnie Brown's Topology and Groupoids asks the reader to find a subset of $\mathbb{R}$ (with usual topology) which does not have a countable neighborhood base.*
My ideas: The subset must be uncountable, since otherwise the unions of countable neighborhood bases of each point would be countable. My intuition was to use $(0,1) \setminus \mathbb{Q}$. Any neighborhood base of $(0,1) \setminus \mathbb{Q}$ must have elements that exclude each closed (in $\mathbb{R}$) subset of $(0,1) \cap \mathbb{Q}$. For instance, given a sequence of rational numbers $q_n$ which has only rational limit points $L \subset \mathbb{Q}$, our neighborhood base would have to have an element that does not meet $\{q_n : n \geq 0 \} \cup L$. But I was having trouble using this line of thinking to show that such a neighborhood base must be uncountable.

*For completeness, my understanding of a neighborhood base of a set $S$ is a collection $N_\alpha$ of neighborhoods of $S$ (i.e. sets containing an open set containing $S$), such that for any neighborhood $N$ of $S$, there is some $\alpha$ such that $N \supset N_\alpha$.
 A: 
The subset must be uncountable, since otherwise the unions of countable neighborhood bases of each point would be countable.

That's a mistake. If we take for example $S = \mathbb{N}$, you get a neighbourhood base by taking the product of neighbourhood bases of the points, not the union.
And that gives you an example, for every function $f \colon \mathbb{N} \to (0,+\infty)$ you get a neighbourhood
$$V_f = \bigcup_{n\in \mathbb{N}} (n - f(n), n+ f(n)),$$
and you can show that no countable system of neighbourhoods is a neighbourhood basis.
If $\{ U_k : k \in \mathbb{N}\}$ is a countable system of neighbourhoods of $\mathbb{N}$, you can construct a function $f$ such that $V_f \not \supset U_m$ for all $m$. Without loss of generality, suppose that $U_k \subset B_{2^{-k-1}}(\mathbb{N})$ so that the neighbourhoods of the points of $\mathbb{N}$ are disjoint. Then we get a function $f_k \colon \mathbb{N} \to (0,+\infty)$ via
$$f_k(n) = \sup \{ \varepsilon > 0 : (n-\varepsilon, n+\varepsilon) \subset U_k\}$$
and have $V_{f_k} \subset U_k$. Let
$$f(n) = 2^{-n-1}\min \{ f_k(n) : k \leqslant n\}.$$
Then $U_k \not \subset V_f$ for all $k \in \mathbb{N}$.
