There is a slight misunderstanding here about the product (and box) topologies. It is not true that every open set (in either topology) is a product of open sets. Indeed, consider the open unit ball $\{(x,y)\in\mathbb{R}^2 \mid x^2+y^2< 1\}$ in $\mathbb{R}^2$. This set is open in both topologies (because they coincide for finite products), but it is not of the form $U\times V$ with $U,V\subseteq\mathbb{R}$ open.
The key fact is that these products form a basis for the topology.
If $X=\prod_{i\in I}X_i$ (as sets) and $(X_i)_{i\in I}$ are topological spaces, then the sets of the form
$$
\prod_{i\in I}U_i,
$$
where $U_i\subseteq X_i$ is open for each $i$, form a base for the box topology on $X$.
Recall that if $B$ is a base which generates the topology $(X,\tau)$, then
$$
\tau = \{U\subseteq X \mid U\ \text{is a union of elements of $B$}\}.
$$
This is equivalent to saying that $U\subseteq X$ is open iff for every $x\in U$ there exists $V\in B$ such that $x\in V \subseteq U$. (Try justifying why these two are equivalent.)
On the other hand, we can consider a subbase. If $S$ is a subbase which generates the topology $(X,\tau)$, then
$$
\tau = \{U\subseteq X \mid U\ \text{is a union of finite intersections of elements of $S$}\}.
$$
Note also that if $S$ is a subbase, then we can obtain a base via
$$
B = \{V\subseteq X \mid V\ \text{is a finite intersection of elements of $S$}\}.
$$
This gets to the heart of a matter. We want to prove:
If $X=\prod_{i\in I}X_i$ (as sets) and $(X_i)_{i\in I}$ are topological spaces, then the sets of the form
$$
\prod_{i\in I}U_i,
$$
where $U_i\subseteq X_i$ is open for each $i$ and $U_i=X_i$ for all but finitely many $i$, form a base for the product topology on $X$.
We can do this with the process described above to turn a subbase into a base. We know that the set
$$
S = \{\pi_i^{-1}(U_i) \mid i\in I,\, U_i\subseteq X_i\ \text{is open}\}
$$
is a subbase for the product topology. This means that sets of the form
$$
\pi_{i_1}^{-1}(U_{i_1})\cap \cdots \cap \pi_{i_n}^{-1}(U_{i_n}),
$$
where $i_1,\ldots,i_n\in I$ and $U_{i_k}\subseteq X_{i_k}$ is open for each $k=1,\ldots,n$, forms a base for the topology.
Since the preimage respects unions and intersections, we can assume that $i_1,\ldots,i_n$ are distinct.
Then we have (why?)
$$
\pi_{i_1}^{-1}(U_{i_1})\cap \cdots \cap \pi_{i_n}^{-1}(U_{i_n})
= \prod_{i\in I}U_i,
$$
where
$$
U_i = \begin{cases}
U_{i_k} & \text{if $i=i_k$ for some $k=1,\ldots,n$},\\
X_i & \text{otherwise}.
\end{cases}
$$
In general, definitions containing "$\dots$" means that everybody can intuitively understand them, but they don't provide details for a formal and rigorous definition of the object.
What is the problem? If you want to prove some properties about an object defined by means of "$\dots$", since you don't have a rigorous definition, you don't know exactly how to formally prove it, even though you intuitively understand what you have to prove: your proof will be inevitably hand-waved, and this could be a way to overlook some important and unexpected details (for instance, the fact that you need some further hypothesis to prove the desired property).
So, your definition of the set of positive integers as $\mathbb{Z}_+= \{1, 1+ 1, 1+ 1 + 1, \dots\}$ is perfectly understandable but then if you want to prove something about $\mathbb{Z}_+$, what do you do? Thanks to the rigorous and formal definition of $\mathbb{Z}_+$ as the smallest inductive subset of $\mathbb{R}$, it is clear which are the elements of $\mathbb{Z}_+$ and how you can use and refer to them.
By the way, there are many different but equivalent ways to define $\mathbb{Z}_+$ formally and rigorously. As @Théophile pointed out in his witty comment,
topologists love unions and intersections
so this is a possible reason why, in his topology handbook, Munkres defined $\mathbb{Z}_+$ as the intersection of some subsets of $\mathbb{R}$. This is not only a joke, but also due to the fact that this definition (among all the possible ones) is maybe the handiest one to deal with positive integers in a topological context, where you usually cope with intersections and unions.
Best Answer
Surjectivity is needed in order that every element of the set $\mathcal A$ is associated to a label, which is something that we want.