This is a two part question. The first part, part (i), I present with the solution I reached. The second part, part (ii) is where I need help.
(i) Let $B$ be a basis for a topology $T$ on a non-empty set $X$. If $B_1$ is a collection of subsets of $X$ such that $T\supseteq B_1\supseteq B$, prove that $B_1$ is also a basis for $T$.
Let $U$ denote any open set in the topology. Since $B$ is a basis for the topology, we know that any open set $U=\bigcup_{i\in j} B_j, B_j \in B$ for all j,
But since $B_1 \supseteq B$, we know that $B_j \in B_1$ for all j, and so thus: $U=\bigcup_{i\in j} B_j, B_j \in B_1$ for all j. Any open set in the topology is therefore a union of members of $B_1$ showing that $B_1$ is indeed a basis for the topology.
So far, I'm ok. But then this I have trouble with:
(ii) Deduce from (i) that there exists an uncountable number of distinct bases for the Euclidean topology on $\mathbb{R}$
Please dumb down the explanation as much as possible. Thanks.
HINT: Let $\mathscr{B}$ be the base of open intervals with rational endpoints. Note that for each $x\in\Bbb R$ the set $U_x=\Bbb R\setminus\{x\}$ is open and not in $\mathscr{B}$. Now use your first result to get uncountably many different bases for the usual topology on $\Bbb R$.