Preamble
I believe that the OP is seeking a characterization of $ \mathbb{Q} $ using only the first-order language of fields, $ \mathcal{L}_{\text{Field}} $. Restricting ourselves to this language, we can try to uncover new axioms, in addition to the usual field axioms (i.e., those that relate to the associativity and commutativity of addition and multiplication, the distributivity of multiplication over addition, the behavior of the zero and identity elements, and the existence of a multiplicative inverse for each non-zero element), that describe $ \mathbb{Q} $ uniquely.
Any attempt to describe the smallest field satisfying a given property must prescribe a method of comparing one field with another (namely using field homomorphisms, which are injective if not trivial), but such a method clearly cannot be formalized using $ \mathcal{L}_{\text{Field}} $.
1. There Exists No First-Order Characterization of $ \mathbb{Q} $
The answer is ‘no’, if one is seeking a first-order characterization of $ \mathbb{Q} $. This follows from the Upward Löwenheim-Skolem Theorem, which is a classical tool in logic and model theory.
Observe that $ \mathbb{Q} $ is an infinite $ \mathcal{L}_{\text{Field}} $-structure of cardinality $ \aleph_{0} $. The Upward Löwenheim-Skolem Theorem then says that there exists an $ \mathcal{L}_{\text{Field}} $-structure (i.e., a field) $ \mathbb{F} $ of cardinality $ \aleph_{1} $ that is an elementary extension of $ \mathbb{Q} $. By definition, this means that $ \mathbb{Q} $ and $ \mathbb{F} $ satisfy the same set of $ \mathcal{L}_{\text{Field}} $-sentences, so we cannot use first-order logic to distinguish $ \mathbb{Q} $ and $ \mathbb{F} $. In other words, as far as first-order logic can tell, these two fields are identical (an analogy may be found in point-set topology, where two distinct points of a non-$ T_{0} $ topological space can be topologically indistinguishable). However, $ \mathbb{Q} $ and $ \mathbb{F} $ have different cardinalities, so they are not isomorphic. This phenomenon is ultimately due to the fact that the notion of cardinality cannot be formalized using $ \mathcal{L}_{\text{Field}} $. Therefore, any difference between the two fields can only be seen externally, outside of first-order logic.
2. Finding a Second-Order Characterization of $ \mathbb{Q} $
This part is inspired by lhf's answer below, which I believe deserves more credit. We start by formalizing the notion of proper subfield using second-order logic.
Let $ P $ be a variable for unary predicates. Consider the following six formulas:
\begin{align}
\Phi^{P}_{1} &\stackrel{\text{def}}{\equiv} (\exists x) \neg P(x); \\
\Phi^{P}_{2} &\stackrel{\text{def}}{\equiv} P(0); \\
\Phi^{P}_{3} &\stackrel{\text{def}}{\equiv} P(1); \\
\Phi^{P}_{4} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x + y)); \\
\Phi^{P}_{5} &\stackrel{\text{def}}{\equiv} (\forall x)(\forall y)((P(x) \land P(y)) \rightarrow P(x \cdot y)); \\
\Phi^{P}_{6} &\stackrel{\text{def}}{\equiv} (\forall x)((P(x) \land \neg (x = 0)) \rightarrow (\exists y)(P(y) \land (x \cdot y = 1))).
\end{align}
What $ \Phi^{P}_{1},\ldots,\Phi^{P}_{6} $ are saying is that the set of all elements of the domain of discourse that satisfy the predicate $ P $ forms a proper subfield of the domain. The domain itself will be a field if we impose upon it the first-order field axioms. Hence,
$$
\{ \text{First-order field axioms} \} \cup \{ \text{First-order axioms defining characteristic $ 0 $} \} \cup \{ \neg (\exists P)(\Phi^{P}_{1} ~ \land ~ \Phi^{P}_{2} ~ \land ~ \Phi^{P}_{3} ~ \land ~ \Phi^{P}_{4} ~ \land ~ \Phi^{P}_{5} ~ \land ~ \Phi^{P}_{6}) \}
$$
is a set of first- and second-order axioms that characterizes $ \mathbb{Q} $ uniquely because of the following two reasons:
Up to isomorphism, $ \mathbb{Q} $ is the only field with characteristic $ 0 $ that contains no proper subfield.
If $ \mathbb{F} \ncong \mathbb{Q} $ is a field with characteristic $ 0 $, then $ \mathbb{F} $ does not model this set of axioms. Otherwise, interpreting “$ P(x) $” as “$ x \in \mathbb{Q}_{\mathbb{F}} $” yields a contradiction, where $ \mathbb{Q}_{\mathbb{F}} $ is the copy of $ \mathbb{Q} $ sitting inside $ \mathbb{F} $.
Your proof does not work. Indeed, subtracting $1$ from $\frac p q$ will give you a rational number, but it will be negative by assumption, so this doesn't help you (since it doesn't give you a contradiction).
A simpler approach: Explicitly state what the infinitely-many positive rationals less than $x$ are.
Hint: If $y$ is a positive rational, what can you say about $\frac{y}2$? About $\frac{y}4$? $\frac{y}8$? ...
Best Answer
If you read this question then any $2$-maximal field $\mathbb M$ contains every square root. But such a field is not the real algebraic numbers since it contains no elements of order $3$ such as $\sqrt[3]{2}$. Of course such a field is probably larger than you want. Instead let $a_n$ be an enumeration of the positive integers (or positive primes) and let $L_0=\mathbb Q$, $L_i=L_{i-1}(\sqrt{a_i})$ then
$$L=\bigcup_{i=0}^\infty L_i$$ is your desired field. Notice that $[L_i:L_{i-1}]=1,2$. Every element of $L$ is necessarily has a power of $2$ since each for each finite $n$, we have that $L_n$ is a power of $2$ extension of $L$. We also have that $L$ is not $2$-maximal because $x^2+1$ does not split over it. Notice that $L$ is an abelian extension as well, since it is generated by its degree $2$ subfields which are necessarily abelian and contained in $\mathbb Q^{\mathrm{ab}}$. So its Galois group should be
$$\lim_{\overleftarrow{n \in \mathbb N}} \left(\mathbb Z_2\right)^n.$$ I've never been that good at inverse limits but I imagine that this is $\prod_{n \in \mathbb N} \mathbb Z_2$, if someone could confirm this I would appreciate it.