I would like some feedback or any tips to improve my proof-writing.
If $A,B\subseteq \mathbb{R}$ are open and dense subsets in $\mathbb{R}$, then $A\cap B$ is open and dense in $\mathbb{R}$.
Proof.
We consider $c\in A\cap B \iff c\in A\land c\in B$. As $c\in A$, then because $A$ is open, there exists $\varepsilon_A>0$ such that $c\in (c-\varepsilon_A,c+\varepsilon_A)\subset A$, and as $c\in B$, then there exists $\varepsilon_B$ such that $c\in (c-\varepsilon_B,c+\varepsilon_B)\subset B$. We take $\alpha = \max\{c-\varepsilon_A,c-\varepsilon_B\}$ and $\beta=\min\{c+\varepsilon_A,c+\varepsilon_B\}$, and see that $c\in (\alpha,\beta)=(c-\varepsilon_A,c+\varepsilon_A)\cap (c-\varepsilon_B,c+\varepsilon_B)\subset A\cap B$. Therefore, whatever may be $c \in A\cap B$, it is an interior point of $A\cap B$, and therefore $A\cap B$ is open.
Again, as $A$ and $B$ are open, there exists $\varepsilon_A>0$ such that $a\in (a-\varepsilon_A,a+\varepsilon_A)\subset A$ for all $a\in A$, and also exists $\varepsilon_B$ such that $b\in (b-\varepsilon_B,b+\varepsilon_B)\subset B$ for all $b\in B$. Well, but the subsets $A,B$ are, not only open, but also dense, then we can expand this formulation such that whatever may be $x\in \mathbb{R}$ and $\varepsilon_A>0$, there will be some $a\in A$ such that $a\in (x-\varepsilon,x+\varepsilon)$, and the same argument can be applied to the subset open and dense $B$. We see, therefore, that whatever may be $x\in\mathbb{R}$ and $\varepsilon>0$, we have that $(x-\varepsilon_A,x+\varepsilon_A)\cap A\neq \varnothing$. We then verify that for whatever value for $x\in\mathbb{R}$ and $\varepsilon>0$ we have $(x-\varepsilon_B,x+\varepsilon_B)\cap B \neq \varnothing$ because $B$ is also dense. But if $A$ is dense, then exists $a\in A$ such that $a\in (x-\varepsilon_B,x+\varepsilon_B)\cap B$ and if $B$ is dense then exists $b\in B$ such that $b\in(x-\varepsilon_A,x+\varepsilon_A)\cap A$. We take $c=a=b$, and therefore $c\in A$ and $c\in B$, we see that exists $c\in A\cap B$ such that $c\in (x-\varepsilon_A,x+\varepsilon_A)\cap (x-\varepsilon_B,x+\varepsilon_B)\cap A\cap B\neq \varnothing$. As this is true for all $x\in \mathbb{R}$, it follows that $A\cap B$ is dense.
Best Answer
Your first paragraph is correct but almost certainly unnecessary, as you surely already know that the intersection of two open sets is always open. There are several problems with your second paragraph, however.
First, your notation implies that there is a single $\epsilon_A>0$ such that for each $a\in A$, $(a-\epsilon_A,a+\epsilon_A)\subseteq A$; this, however, is false. For each $a\in A$ there is an $\epsilon_a>0$ such that $(a-\epsilon_a,a+\epsilon_a)\subseteq A$, but $\epsilon_a$ depends on $a$. The same comments apply, mutatis mutandis, to $\epsilon_B$. This could be bad writing on your part rather than a failure of understanding, but there is a clear error later in the paragraph when you say that we can take $c=a=b$: at that point in the argument you have no reason to think that it is possible to choose $a$ and $b$ to be the same point. Indeed, that is essentially what you are trying to prove.
Specifically, you want to prove that for any $x\in\Bbb R$ and any $\epsilon>0$,
$$(x-\epsilon,x+\epsilon)\cap(A\cap B)\ne\varnothing\,.$$
$A$ is dense in $\Bbb R$, so we know that $$(x-\epsilon,x+\epsilon)\cap A\ne\varnothing\,;$$ let $a\in(x-\epsilon,x+\epsilon)\cap A$. Since $(x-\epsilon,x+\epsilon)\cap A$ is open there is a $\delta>0$ such that
$$(a-\delta,a+\delta)\subseteq(x-\epsilon,x+\epsilon)\cap A\;.$$
And $B$ is dense in $\Bbb R$, so we know that
$$(a-\delta,a+\delta)\cap B\ne\varnothing\,.$$
Let $b\in(a-\delta,a+\delta)\cap B$. Then
$$\begin{align*} b&\in(a-\delta,a+\delta)\cap B\\ &\subseteq\big((x-\epsilon,x+\epsilon)\cap A\big)\cap B\\ &=(x-\epsilon,x+\epsilon)\cap(A\cap B)\,, \end{align*}$$
i.e., $(x-\epsilon,x+\epsilon)\cap(A\cap B)\ne\varnothing$, and therefore $A\cap B$ is dense in $\Bbb R$.