Understanding the choice function Let $X = \{A_1, A_2, A_3,\ldots, A_k\}$ where $A_i$ is a non-empty set. Then for every $A_i, f(A_i) \in A_i$. Is that the right definition?
Example: Let $B_1 = \{1, 2\}$  and  $Y = \{B_1\}$. Consider $f(x) = 2x$. Then $f(B_1) = \{2, 4\}$. Does it mean $\{2, 4\} \in \{1, 2\}$?
 A: In your case, $f$ is not a choice function. A valid choice function $f:Y \to \bigcup_i B_i$ would be given either by $f(B_1) = 1$ or $f(B_1) = 2$. In that case we would have $f(B_1) \in B_1$.
For a bigger example, let $Y = \{\Bbb R, \Bbb C, [2, 7), \{1, 2, 3\}\}$. In that case, the function $f$ that always gives the answer $3$ is a valid choice function, but so is the function $g$ given by
$$
g(\Bbb R) = \pi \quad (\in \Bbb R)\\
g(\Bbb C) = \ln(2) + \sqrt2i\quad (\in \Bbb C)\\
g([2, 7)) = e \quad(\in [2, 7))\\
g(\{1, 2, 3\}) = 1 \quad(\in\{1, 2, 3\})
$$
Also note that while in this case, the elements of $Y$ are all subsets of $\Bbb C$, that need not be the case. The elements of $Y$ can be anything non-empty. There needn't be any connection between the different $B_i$. And that is what makes the existence of choice functions for general infinite sets $Y$ into such a strong statement.
And to adress your question in the comments, in this setting it is common to thing of everything as sets. Numbers are sets, functions are sets, relations are sets. The most common construction of the natural numbers, for instance, is to let $0$ be the empty set $\{\}$, then let $n+1$ be given by $n\cup \{n\}$. That means that $1 = \{\{\}\}$ and $2 = \{\{\}, \{\{\}\}\}$
