The word "infinite" is unnecessary.
But it is true that $A$ is a union (possibly infinite, but that is irrelevant) of open sets, hence is open.
I'll try to avoid what seems like a use of AC (for each $x$, pick a $B_x$ with the appropriate properties). For each $x\in A$, let $\mathcal{B}_x = \{B\subseteq X\mid B\text{ is open, }x\in B,\text{ and }B\subseteq A\}.$ We know, by assumption, that $x\in\bigcup\mathcal{B}_x$ (since the set is not empty). (Recall that if $X$ is a set whose elements are sets, then $\bigcup X$ is the union of the elements of $X$). Moreover, $\bigcup\mathcal{B}_x$ is a union of open sets, hence open, and every element of $\mathcal{B}_x$ is contained in $A$, hence $\bigcup\mathcal{B}_x$ is contained in $A$.
Now consider $B = \bigcup_{x\in A}\left(\bigcup\mathcal{B}_x\right).$ This is open, being a union of open sets; contained in $A$, being a union of sets contained in $A$; and contains $A$, since for each $x\in A$, $x\in\bigcup\mathcal{B}_x\subseteq B$. Thus, $A=B$, so $A$ is open.