Following this definition (Shilling - Measures, Integrals and Martingales):
Let $f: X \to X'$ be a map and let $\mathcal{A'}$ be a $\sigma$-algebra in $X'$. Then:
$$\mathcal{A}:= f^{-1}(\mathcal{A'}) :=\{f^{-1}(A'): A' \in \mathcal{A'}\}$$
is a $\sigma$-algebra in $X$.
Now, I probably do something very illegal here but let's consider the following mapping:
$X = \{1, 2\}$, $X' = \{D, \emptyset \}$ where $D$ is arbitrary set.
$\{1\} \mapsto \{{D}\}$
$\{2\} \mapsto \{{\emptyset}\}$
$X'$ is a minimal $\sigma$-algebra in $D$. But it seems $X$ is not, since we do not have mapping from union of $\{1\} \cup\{2\}$ to anything, so $X$ is not closed under countable unions. How is this "counterexample" wrong?