Let $A_n$ be such a base. For each $A_n\subset A_m$ pick, if possible, an interval $I_{n,m}:=[a_{n,m},b_{n,m})$ such that $A_n\subset [a,b)\subset A_m$.
The collection $I_{n,m}$ should be a base. In fact, if $A$ is open, take and arbitrary point $x\in A$. Then there is (because the $A_n$ for a base) an $A_n\ni x$ contained in $A$. There is (because the $[a,b)$ form a base) a small $[a,b)\ni x$ contained in $A_n$ and there is (because $A_n$'s are a base) an $A_m\ni x$ contained in $[a,b)$. Therefore there is an $I_{n,m}$ (because for this particular $n,m$ is is possible such $I_{n,m}$. Notice that $[a,b)$ could be a candidate). This $I_{n,m}$ incidentally contains $x$ and is inside $A$. Therefore $A$ is equal to the union of all the $I_{n,m}$ inside it.
Above we didn't really used anything about the particular bases. In general: Given two bases you can construct a subset of one, that is still a basis, and has cardinality not larger than the other base.
Therefore there is a countable base $I_{n,m}=[a_{n,m},b_{n,m})$. Since $\mathbb{R}$ is uncountable, there are two points $x,y$ such that they are not boundaries of any $I_{n,m}$.
But $[x,y)$ can't be formed by a union of $I_{n,m}$, the point $x$ is never covered by the $I_{n,m}$ lying inside the interval $[x,y)$.