So I was working on showing that
$$\operatorname{Th}(\Bbb{R}, 0,1,+, \le) = \operatorname{Th}(\Bbb{Q}, 0,1,+, \le) $$
My initial idea for working on this problem was to systematically start by showing $$\operatorname{Th}(\Bbb{R}, 0,1,+) = \operatorname{Th}(\Bbb{Q}, 0,1,+) $$ and $$\operatorname{Th}(\Bbb{R}, 0,1,\le) = \operatorname{Th}(\Bbb{Q}, 0,1,\le).$$
Hopefully by then I would have enough intuition of the problem to tackle the larger case. But it occurred to me I don't understand enough from these cases. Starting for the first one. I noted that since there are no actual relations here, and just operators, thus there is no way to set up an equation that only has a solution in one but not the other. Or more generally, its impossible to make statements that have any truth or false value. Other than just statements involving existence. (If we use foralls, given the non existence of other formula, they don't create any constraint)
For the next one it's clear that these are equivalent since no operators can be used so only expressions involving comparing the $1$ and $0$ (present in both) can be manufactured and since we cannot access the underlying structure in either collection of expressions there's no way to differentiate our structures here.
So given that both cases are "obvious" when we mix it up and now consider the full:
$$\operatorname{Th}(\Bbb{R}, 0,1,+, \le) = \operatorname{Th}(\Bbb{Q}, 0,1,+, \le) $$
One can observe that $+$ is closed in $\Bbb{Q}$ and that the rationals are dense over the reals. But that doesn't say much since $\times$ is also closed in the rationals yet its trivial to show:
$$\operatorname{Th}(\Bbb{R}, 0,1,+, \times, \le) \ne \operatorname{Th}(\Bbb{Q}, 0,1,+,\times \le) $$
Namely consider
$$ \exists x \forall y |1 + 1 \le x*x \le 1 + 1 $$
This sentence is only true for $\Bbb{R}$ as the $x$ in question that makes it true does not exist in $\Bbb{Q}$.
So my earlier intuition about closure and density have no substantial weight to them.
But then how do I proceed further?
I recommend using Ehrenfeucht-Fraisse games (this is possibly because I just really like them :P).
Given a pair of relational (see below) structures $\mathcal{A}$ and $\mathcal{B}$ and a natural number $n$, consider the following game $EF_n(\mathcal{A},\mathcal{B})$ between two players, Mix and Match:
On turn $k$, first Mix plays an element of one structure or the other: either $a_k\in\mathcal{A}$ or $b_k\in\mathcal{B}$. Then Match plays an element of the opposing structure: either $b_k\in\mathcal{B}$ or $a_k\in\mathcal{A}$.
Play continues for $n$ turns - at the end of which, Mix and Match have built a sequence $a_1, a_2, . . . , a_n$ of elements of $\mathcal{A}$ and a sequence $b_1, b_2, . . . , b_n$ of elements of $\mathcal{B}$.
Now, we look at the atomic diagrams of the tuples $(a_i)$ and $(b_i)$: if there is a quantifier-free formula $\varphi(x_1, . . . , x_n)$ such that $$\mathcal{A}\models\varphi(a_1, . . . , a_n)\quad\mbox{ but }\quad\mathcal{B}\not\models\varphi(b_1, . . . , b_n),$$ then Mix wins; otherwise (that is, if the tuples "look the same") Match wins.
Here's the remarkable fact:
So you can prove the elementary equivalence of two structures by describing a winning strategy for Match (and, of course, showing that it actually is a winning strategy).
EF-games are definitely a sometimes technique: other methods are often much faster. However, personally I find EF-games the most intuitively satisfying approach, and I don't really understand why two structures are elementarily equivalent until I know how to win the EF-game between them.
OK, what about that "relational" business? Well, a relational structure is just a structure which only has constants and relations - no functions. Bad news: you've got functions! Good news: you can get rid of them! We can "relationalize" a structure $\mathcal{A}$ by replacing each $n$-ary function symbol $f$ with an $(n+1)$-ary relation symbol $G_f$, the graph of $f$: $$G_f^\mathcal{A}=\{(a_1, . . . , a_n, a_{n+1}): f^{\mathcal{A}}(a_1, . . . , a_n)=a_{n+1}\}.$$
To see why relationalization is necessary, imagine we played the length-1 EF-game with the functional structures $\mathcal{A}=(\mathbb{R}, +, 1)$ and $\mathcal{B}=(\mathbb{Q}, +, 1)$. As Mix, I would play $a_1=\pi\in\mathcal{A}$; Match would have to play some rational $b_1\in \mathcal{B}$. Suppose $b_1={p\over q}$; then $b_1$ would satisfy the atomic formula $b_1+. . . .+b_1=1+. . . .+1$ ($q$-many "$+$"s on the left, $p$-many on the right) in $\mathcal{A}$. But of course $\pi$ doesn't satisfy the corresponding formula in $\mathcal{A}$. So I, Mix, would win!
On the relationalization, by contrast, I wouldn't be able to pull that trick - it would take me $\max\{q, p\}$-many moves to be able to "show" that $b_1$ and $\pi$ were different. So to match $\pi$ in the game of length $n$, I would just need to pick a $b_1$ which is "close to $\pi$ and not too rational" - for instance, the nearest rational with denominator $n+1$. If you try playing, say, $EF_2$ of the two structures, you'll start to see what a winning strategy for Match looks like . . .
A similar idea - that we can find "definable" elements which "look like" undefinable elements for long enough to win specific EF-games - works for showing that the linear orders $\mathbb{Z}$ and $\mathbb{Z}\oplus\mathbb{Z}$ are elementarily equivalent; see chapter 6 of Rosenstein's wonderful book "Linear Orderings" (https://books.google.com/books?id=y3YpdW-sbFsC&pg=PA93&lpg=PA93&dq=ehrenfeucht-fraisse+game+Z+linear+order&source=bl&ots=eREBRBKTDL&sig=9aBq9q7YLHzB6ujtpeW92IPoFK4&hl=en&sa=X&ved=0CCAQ6AEwATgKahUKEwj6zq3ewpTJAhUD02MKHRTwAnQ#v=onepage&q=ehrenfeucht-fraisse%20game%20Z%20linear%20order&f=false).