[Incorrect answer, perhaps mendable, perhaps not, sorry, see my comments below]

The answer is yes, the statement is in fact equivalent to AC.

AC is equivalent to "every set of pairwise disjoint nonempty sets has a choice set". So to see that the statement implies AC we can let ${\cal A}\subseteq{\cal P}(X)$ consist of pairwise disjoint sets.

Let $M\subseteq X$ be a *minimal cover* for ${\cal A}$ , that is: for all $m\in M$ the set $M\setminus \{m\}$ is not a cover of ${\cal A}$. Then $M$ is a choice set for ${\cal A}$.
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To see that AC implies the statement, we use Zorn's lemma.

For given ${\cal A}\subseteq{\cal P}(X)$ with the property that $A\neq B \in {\cal A}$ implies $|A\cap B|\leq 1$, let $H\subset {\cal P}({\cal A})\times{\cal P}(X)$ be defined by:
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$$H := \ \{(K,L)\in {\cal P}({\cal A})\times{\cal P}(X)\ |\ L\ \mbox{is a minimal cover for}\ K\ \}$$

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We define a partial order on $H$ using set-inclusion:
$$(K,L)\leq(K',L'):= K\subseteq K' \wedge L\subseteq L'$$
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For $J=(K,L)\in H$ let $J_0=K, J_1=L$ be the coordinate projections.

We verify that every ascending chain in $(H,<)$ has an upper bound. Let $G\subseteq H$ such that $<$ is a total order on $G$. Then the upper bound $T$ for $G$ is given by:

$$T = (\bigcup_{J\in G} J_0, \bigcup_{J\in G} J_1)$$

It is straightforward to see that $\bigcup_{J\in G} J_1$ is again a minimal cover for $\bigcup_{J\in G} J_0$, so indeed $T$ is in $H$.

It is easy to see that there are non-empty ascending chains. By Zorn's lemma, $(H,<)$ contains a maximal element $U=(V,W)$. We claim that $V={\cal A}$, or in other words: $W$ is the desired minimal cover for ${\cal A}$.

To see that $V={\cal A}$, suppose there is $X\in{\cal A}, X\not\in V$.

case 1) $X\cap W \not= \emptyset$. Then $(V\cup\{X\},W)$ is larger than $(V,W)$, contradiction.

case 2) $X\cap W = \emptyset$. Pick $x\in X$. Then $(V\cup\{X\},W\cup\{x\})$ is larger than $(V,W)$, again contradiction.