Let $I \subseteq \mathbf{R} $ be dense and open.
Because $I$ is dense, for $x \in \mathbf{R}$, $$\exists y\in I:y=I \cap B_{\epsilon}(x)$$
Because $I$ and $B_{\epsilon}(x)$ are open, $I \cap B_{\epsilon}(x)$ is also open and thus $\exists \delta, \text{as a function of y, }\delta(y) \text{, such that}$: $$B_{\delta}(y) \subseteq I \cap B_{\epsilon}(x) \tag{*}$$
Question: Is there a way to construct $y$ so that we can have an upper bound for $\delta (y)$ that makes (*) holds? In other words, can we find a $y' \in I$ such that:
$$\delta (y') < \alpha \to B_{\delta(y')}(y')\subseteq I \cap B_{\epsilon}(x)$$
Let $\left( X, d \right)$ be any metric space with the $d$ as the distance function. Consider your dense open subset $I$ and any point $x \in X$. Now for a given epsilon, consider $I \cap B_{\epsilon} \left( x \right)$. Now, if $y$ belongs to the intersection, in particular, $y$ belongs to $I$.
Therefore, $\exists \delta_1 > 0$ such that $B_{\delta_1} \left( y \right) \subseteq I$.
Also, $y$ is in $B_{\epsilon} \left( x \right)$. Therefore, $\exists \delta_2 > 0$ such that $B_{\delta_2} \left( y \right) \subseteq B_{\epsilon} \left( x \right)$. Now, if you choose $\delta = \min \left\lbrace \delta_1, \epsilon - d \left( x, y \right) \right\rbrace$, then $B_{\delta} \left( y \right) \subseteq I \cap B_{\epsilon} \left( x \right)$ which will do your work.
I have attached an image below for better understanding. In the image, $y$ is shown with $2$ open balls, one is relative to $I$ and other is relative to $B_{\epsilon} \left( x \right)$. Once you choose the lesser radius, you are done with what you want.