I am reading Measure and Category by John Oxtoby, and I have a question about one of the proofs in the book. I reproduce the proof here in full for convenience.
Theorem 7.3. If $f$ can be represented as the limit of an everywhere convergent sequence of continuous functions, then $f$ is continuous except at a set of points of first category.
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Proof. It suffices to show that, for each $\varepsilon>0$, the set $F=\{x: \omega(x) \geqq 5 \varepsilon\}$ is nowhere dense. Let $f(x)=\lim f_n(x), f_n$ continuous, and define $$ E_n=\bigcap_{i, j \geqq n}\left\{x:\left|f_i(x)-f_j(x)\right| \leqq \varepsilon\right\} \quad(n=1,2, \ldots) . $$ Then $E_n$ is closed, $E_n \subset E_{n+1}$, and $\bigcup E_n$ is the whole line. Consider any closed interval $I$. Since $I=\bigcup\left(E_n \cap I\right)$, the sets $E_n \cap I$ cannot all be nowhere dense. Hence, for some positive integer $n, E_n \cap I$ contains an open interval $J$. [Emphasis mine.] We have $\left|f_i(x)-f_j(x)\right| \leqq \varepsilon$ for all $x$ in $J ; i, j \geqq n$. Putting $j=n$ and letting $i \rightarrow \infty$, it follows that $\left|f(x)-f_n(x)\right| \leqq \varepsilon$ for all $x$ in $J$. For any $x_0$ in $J$ there is a neighborhood $I\left(x_0\right) \subset J$ such that $\left|f_n(x)-f_n\left(x_0\right)\right| \leqq \varepsilon$ for all $x$ in $I\left(x_0\right)$. Hence $\left|f(x)-f_n\left(x_0\right)\right| \leqq 2 \varepsilon$ for all $x$ in $I\left(x_0\right)$. Therefore $\omega\left(x_0\right) \leqq 4 \varepsilon$, and so no point of $J$ belongs to $F$. Thus for every closed interval $I$ there is an open interval $J \subset I-F$. This shows that $F$ is nowhere dense.
How does the step in bold follow from the fact that there exists some $n$ such that $E_n \cap I$ is not nowhere dense in $I$? Could we not take a countable sequence $a_{i}$ of real numbers linearly independent over $\mathbb{Q}$, and let $E_i = Q + a_i$ for $i>1$, and $E_1 = \mathbb{R}-\cup_{i } E_i$?