This is a problem in Lee's Introduction to Smooth Manifolds.
Show that a disjoint union of uncountably many copies of $\mathbb{R}$ is not second-countable.
Let $S$ be the disjoint union of uncountably many copies of $\mathbb{R}$. It suffices to construct an open subset $U\subset S$ that cannot be written as the union of countably many open sets in $S$. Somehow, I reckon that one can obtain a contradiction by forming a bijection between the index set of the disjoint union (which is uncountable) and the claimed basis (which is countable).
I try reasoning as follows. Let $\mathcal{B}=\{B_1,B_2,\dots\}$ be a claimed countable basis. Let $A$ be the index set of $S$. Let $A_i\subset A$ be the set of indices $\alpha\in A$ such that there exists an element of the form $(x,\alpha)\in B_i$. Observe that if each $A_i$ was countable, then $A'=\bigcup_{i=1}^{\infty}{A_i}$ is countable as a countable union of countable sets. Since $A$ is uncountable, we must have that $A'$ is a proper subset of $A$. So there exists $\alpha_0\in A\setminus A'$, and now the set $$\{(x,\alpha_0)\colon x\in T\}$$ where $T$ is open in $\mathbb{R}$, is an open set in $S$ that cannot be written using the basis $\mathcal{B}$, a contradiction. Hence, there exists an integer $n$ such that $A_n$ is uncountable.
But I am not sure where to go from here.
The copies of $\mathbb R $ themselves form a collection of open sets that can't be written as the union of countably many open sets. For they form a disjoint collection of cardinality $\ge\mathfrak c $.
To pan it out, each copy of $\mathbb R $ would have to contain an element of the base. But that makes uncountably many (elements of the base).