Is usage of axiom of countable choice unavoidable to prove this?

No, there are no equivalences here: your statement doesn’t imply countable choice, and your statement is not a theorem of ZF.

You don’t need the full power of countable choice. With the notation of your proof, if you are merely given a sequence of non-empty countable subsets $\mathcal E_n\subset \{E:D_n=f^{-1}(E)\}$ then you can define $E_n=\bigcap_{E\in \mathcal E_n} E.$ So you can use the following weaker axiom described in Howard-Rubin’s “Consequences of the Axiom of Choice”:

FORM 131. $MC_\omega(\aleph_0,\infty)$: For every denumerable family $X$ of pairwise disjoint non-empty sets, there is a function $f$ such that for each $x\in X,$ $f(x)$ is a non-empty countable subset of $x.$

Countable choice is “Form 8” and is strictly weaker.

To show that your statement is not a theorem of ZF I will make use of a suitable permutation model.

Work in ZFC for now. One ingredient we need is a poset $P$ such that:

• For every countable subset $B\subset P$ there exists $a,c\in P$ such that $a<b<c$ for all $b\in B.$
• For every $a,b\in P$ there is an order automorphism of $P$ that takes $a$ to $b$

For example take $P$ to be the long line with its usual total order. Let $P^*$ denote the poset obtained from $P$ by adjoining a bottom element $-\infty$ and top element $\infty.$ Let $\mathcal I$ be the family of sets $E\subseteq P^*$ that are determined by a countable sequence of comparisons, in the specific sense that there exists a sequence $b:\omega\to P$ and a set $F\subset\mathcal P(\omega)$ such that $x\in E\iff \{i\in\omega:x \leq b(i)\}\in F.$ This is easily checked to be a $\sigma$-algebra, invariant under order automorphisms of $P^*,$ and has the following property:

$(\dagger)$ If $E\in\mathcal I$ and $E$ is invariant under order automorphisms of $P,$ then $E=\emptyset$ or $E=P^*.$

Proof of $(\dagger)$: For all $a,b\in P$ we have $a\in E\iff b\in E$ by the invariance under an automorphism taking $a$ to $b.$ By definition $E$ is determined by some sequence $p:\omega\to P.$ There exists $a\in P$ with $a\leq b(i)$ for all $i$; since $-\infty\leq b(i)$ as well, we have $-\infty\in E\iff a\in E.$ Similarly there exists $c\in P$ with $c>b(i)$ for all $i,$ so $\infty\in E\iff c\in E.$ Putting this together, if $E\cap P$ is empty so is $E,$ and if $E\cap P$ is nonempty then $E\cap P=P$ and hence $E=P^*.$ $\square$

Take $Y=\omega\times P^*.$ This is the set of atoms for the permutation model. The group of automorphisms is the direct product of countably many copies of the order automorphism group of $P^*,$ in symbols $\mathrm{Aut}(P^*,<)^\omega,$ with the normal filter generated by the subgroup of elements whose first $n$ components are trivial. This group acts on $Y$ by $g(i,g_i(p))=(i,p).$ Construct the permutation model of hereditarily symmetric sets with respect to this group action and normal filter.

The resulting permutation model contains $Y$ and the algebra $\mathcal N$ of subsets of $Y$ such that there exists $n$ such that for each $i\geq n,$ the set $Y\cap(\{i\}\times P^*)$ is either $\emptyset$ or $\{i\}\times P^*.$ Internally, $\mathcal N$ is a $\sigma$-algebra. Let $X=\omega\times \{-\infty,\infty\}\subset Y$ and let $f:X\to Y$ be the inclusion map. Then $\mathcal M=\{f^{-1}(E):E\in\mathcal N\}$ is not a $\sigma$-algebra: it contains the sequence of sets $D_n=n\times\{-\infty\},$ but not $\bigcup_n D_n=\omega\times\{-\infty\}.$

You can use the Jech-Sochor embedding theorem to embed this counterexample in a model of ZF.