Proving Hall's Theorem, struggling proving closedness in product topology on the way. trying to verify that i worked out the details on the closedness in product topology correctly, i marked the part where it gets shaky.
From the book "The Banach-Tarski Paradox" by Stan Wagon:

Theorem C.2 (Hall-Rado-Hall Theorem)
Let $(A,B,E)$ be a bipartite graph such that every vertex in $A$ has finite degree. Then the following holds:
$$
\text{There exists a matching of every vertex in}~A
\iff \text{for all }n \in \mathbb{N}~\text{each set of $n$ vertices in A has at least $n$ neighbors.}
$$

Let $N(a)$ denote the set of neighbors of $a \in A.$
$\Rightarrow$ is trivial since if we have found a matching $f: A \to B$ for $A$, that is an injective map such that $f(a) \in N(a)~\text{for all}~a \in A$, then its restriction to any finite subset $H = \{a_1,\ldots, a_n\} \subset A$ must be a matching for $H$, hence we necessarily have $|\bigcup_{k=1}^n N(a_k)| \geq n$ by injectivity.
$\Leftarrow$ The author argues with the Tychonoff theorem in the following way

*

*Consider $B$ equipped with the discrete topology, which means that every subset is open and a subset is compact iff it is finite.

*Let $\prod_{a \in A} B_a$ be the topological space of the set of all functions $f: A \to B$, equipped with the product topology.

*Consider the subspace $Y = \prod_{a \in A} N(a) \subset \prod_{a \in A} B_a$. Since every $N(A) \subset B$ is finite by assumption and hence compact with respect to the discrete topology, we conclude by Tychonoffs Theorem that $Y \subset B^A$ is a compact subspace.

furthermore consider for each finite non-empty $\{h_1, \ldots, h_n\} = H \subset A$ the set $$M(H)=\{f \in B^A : f(a) \in N(a)~\text{for all $a \in A$} \land f_{|H}~\text{is a matching for H}\}.$$
The first condition on $M(H)$ is equivalent to saying $f \in Y=\prod_{a \in A} N(a)$, hence
$$M(H) = \{ f \in Y : f_{|H} \text{ is injective} \}.$$
By assumption $H$ satisfies the Hall Condition and is finite, hence $M(H)$ is non-empty.
(hoping for confirmation/correction on the following characterization of non-injective functions in terms of preimages of projections)
Now we wish to show that $M(H) \subset Y$ is a closed subset. We have
$$Y \setminus M(H) = \{ f \in Y : f_{|H} \text{ is not injective} \},$$
what this means is we find $h_i,h_j \in H, i \neq j: f(h_i)=f(h_j)$, which is the same as saying we find $ i \neq j$ such that $\pi_i f = \pi_j f$ where $\pi_i: \prod_{a \in A} B_a \to B_{h_i}$ is the projection onto the $h_i$ component. This gives us
$$ Y \setminus M(H) = \bigcup_{i=1}^n \bigcup_{j=1 \\ j\neq i}^n \Big(\pi_i^{-1}(N(h_i)) \cap \pi_j^{-1}(N(h_j)) \Big)$$ where each $\pi_i^{-1}(N(h_i))$ is open by definition of the product topology. Hence $M(H)$ is closed.
(The rest is clear but i shall note it for the sake of completeness.)
We can conclude the proof by noting that for non-empty finite $H,H'\subset A$ we have
$$\emptyset \neq M(H \cup H') \subset M(H) \cap M(H'),$$ since $H \cup H'$ is finite and satisfies the Hall Condition by assumption, on the other hand a matching for $H \cup H'$ is obviously a matching for $H$ (resp. $H'$) when restricted to $H$ (resp. $H'$). Thus the family
$$\mathcal{M}=\{M(H) :\text{non-empty finite }H \subset A\}$$ has the finite intersection property, by compactness of $Y$ we find
$$ f \in \bigcap \{M(H) :\text{non-empty finite }H \subset A\} $$ and such an $f$ is clearly a matching for $A$, qed.
 A: So i realized i managed to ask my question about characterizations of open sets (resp. convergence) with respect to the product topology in a very convoluted way. i read up a little and i want to share what i learned.
So we are given a topological space $B$ and the space of functions
$$\prod_{a\in A} B_a =B^A=\{f: A \to B\} \quad \text{where $B_a = B$ for all $a \in A$},$$ endowed with the product topology (i.e. the coarsest topology for which the projections $\pi_a: \prod_{a \in A} B_a \to B_a$ are continuous for all $a \in A$).
In Thm. 11.9 "General Topology" by Willard it is shown that for all nets $(f_i)_{i \in I}$ with $f_i \in \prod_{a \in A} B_a$ the following holds:
$$\lim_{i \in I} f_i = f \iff \lim_{i \in I} \pi_a f_i = \pi_a f \quad \text{for all $\pi_a$, $a \in A$}.$$
So for the purposes of the theorem we consider $B$ equipped with the discrete topology and we consider, for all finite non-empty subsets $H=\{h_1, \ldots, h_n\} \subset A$ the set $$E(H)=\{f \in \prod_{a \in A} B_a : f_{|H} \text{ is injective} \} \subset \prod_{a \in A} B_a.$$
We claim that $E(H)$ is closed. So let $(f_i)_{i \in I}$ be a net in $E(H)$ which converges to some $f \in \prod_{a \in A} B_a.$ Then by the cited Thm. from Willard, we must have $\lim_{i\in I} \pi_a f_i = \pi_a f$ for all $a \in A,$ which claims a nets convergence in a discrete space (which is only possible if it is identical to its limit eventually).
That means for each $h_k \in H$ we find some $i_k$ such that $\pi_{h_k}f_i=f_i(h_k)=f(h_k)=\pi_{h_k}f$ for all $i \geq i_k$. We conclude that $f_{i\ {|H}}=f_{|H}$ for all $i \geq i_o$ where $i_o$ is some common upper bound of $\{i_1, \ldots, i_n\}$.
The claim in my original post follows by noting that $M(H) = E(H) \cap Y$ which is still closed.
unclear: does it suffice to consider sequences rather then nets in this case?
