Consider the following premises in a symbolic expression: $$ A, B \subseteq \mathbb{R}\\ \forall x \in A: \exists y \in B: y \leq x\\ |A| < \infty $$ Then, we can conclude that $\exists y \in B: \forall x \in A : y \leq x $. Note that the order of the bounded quantifiers is exchanged and this is not a first-order consequence.
One of the proofs can be informally stated that for each element of $A$, we can pick an element from $B$ that is not larger than $x$. Since $A$ is a finite set, the smallest element of $B$ is not larger than every element of $A$.
Can this proposition be proved by natural deduction? I tried to show this using rules of inference presented in a Fitch form in vain.
As Mauro said, you did not express "$A$ is finite" in a logical form. So how would it be possible to do anything with it? Suppose you use this Fitch-style system. Then the simplest way to express it would be as follows.
Given any set $A$, define $A$ to be finite iff:
$∃c{∈}ℕ\ ∃f{∈}FN(ℕ_{<c},A)\ ∀x{∈}A\ ∃k{∈}ℕ_{<c}\ ( \ f(k)=x \ )$.
In English, $A$ is finite iff there is some $c{∈}ℕ$ and some surjection from $ℕ_{<c}$ onto $A$. Here $ℕ_{<c}$ is defined as $\{ k : k{∈}ℕ ∧ k<c \}$. If you want to minimize set theory, you can use the following alternative (which is equivalent over set theory).
Given any set $A$, define $A$ to be finite iff:
$∃c{∈}ℕ\ ∃f{∈}FN(ℕ,A)\ ∀x{∈}A\ ∃k{∈}ℕ\ ( \ k<c \land f(k)=x \ )$.
You are correct that you cannot automatically swap the quantifiers. To prove your desired theorem, you would need the induction rule (see under "Set Theory" in the linked post). See if you can find a proof!