Show that this is not a countable cover for the graph of a continuous function. Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function. Consider any rational number $q$. For any $\epsilon >0$, there is a $\delta > 0$ such that $|q - y| < \delta \implies |f(q) - f(y)| < \epsilon$.
I was using this to try and make a countable cover of the graph of $f$ using rectangles of vertical height $2 \epsilon$ and horizontal width $2 \delta$. There are countable number of rationals number and so there are countable number of such rectangles. In any interval that these rectangles might "miss", there is a rational number, and so a covering rectangle.
What is the flaw in this argument? It is possible for these rectangles to miss a part of the graph of $f$?
 A: Yes, it’s possible. Fix $\epsilon>0$. For each $q\in\Bbb Q$ you have a $\delta_q>0$ such that $|f(x)-f(q)|<\epsilon$ whenever $|x-q|<\delta_q$. Let $$R_q=(q-\delta_q,q+\delta_q)\times\big(f(q)-\epsilon,f(q)+\epsilon\big)\;;$$ the open rectangle $R_q$ covers $\{\langle x,f(x)\rangle:|x-q|<\delta_q\}$, so $\{R_q:q\in\Bbb Q\}$ covers the graph of $$f\upharpoonright\bigcup_{q\in\Bbb Q}(q-\delta_q,q+\delta_q)\;,$$ but the intervals $(q-\delta_q,q+\delta_q)$ need not cover $\Bbb R$, so the rectangles $R_q$ need not cover the graph of $f$.
To see why the intervals $(q-\delta_q,q+\delta_q)$ need not cover $\Bbb R$, let $\alpha\in\Bbb R$ be irrational, and suppose that $\delta_q<|q-\alpha|$ for each $q\in\Bbb Q$; then for each $q\in\Bbb Q$ we have $\alpha\notin(q-\delta_q,q+\delta_q)$. In fact it’s possible for 
$$\Bbb R\setminus\bigcup_{q\in\Bbb Q}(q-\delta_q,q+\delta_q)$$
to contain a set homeomorphic to the middle-thirds Cantor set, in which case
$$\left\vert\Bbb R\setminus\bigcup_{q\in\Bbb Q}(q-\delta_q,q+\delta_q)\right\vert=|\Bbb R|=2^\omega=\mathfrak{c}\;,$$
and the rectangles $R_q$ fail to cover a lot of points of the graph of $f$.
