Here, "by induction" means define the sets inductively. That is
1) Show that the open set $U_1$ can be constructed so that it meets $A$.
2) Assume that the open sets $U_1$, $U_2$, $\ldots\,$, $U_n$, have been constructed so that each $U_i$ meets $A$ and that $\overline {U_i}\cap \overline{U_j}=\emptyset$ whenever $i\ne j$ and $1\le i,j\le n$.
3) Then show that $U_{n+1}$ can be constructed to have the appropriate properties.
In order to construct the required infinite sequence of open sets, you will need to impose the additional property that for each $j$ the complement of $\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_j}$ contains infinitely many points of $A$.
(Without this requirement, the induction may stop after a finite number of steps.)
So, you need to do the following:
1) Show that the open set $U_1$ can be constructed so that it meets $A$ and $\overline{U_1}^C\cap A$ is infinite.
2) Assume that the open sets $U_1$, $U_2$, $\ldots\,$, $U_n$, have been constructed so that
$\ \ \ \ $a) each $U_i$ meets $A$,
$\ \ \ \ $b) $\overline {U_i}\cap \overline{U_j}=\emptyset$ whenever $i\ne j$ and $1\le i,j\le n$,
$\ \ \ \ $c) $ (\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n} )^C\cap A$ is infinite.
3) Then show that an open set $U_{n+1}$ can be constructed so that
it meets $A$, its closure is disjoint from the closure of each previously defined set, and that $ (\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n}\cup\overline{U_{n+1}})^C\cap A$ is infinite.
To establish 1), you could argue as Martin does in his answer.
Assuming that 2) holds, to establish 3), you would use regularity:
$F_n=\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n}$ is closed and by 2c) its complement contains infinitely many points of $A$. Given $a\in F_n^C\cap A$, by regularity, you can find an open set $U_{n+1}$ containing $a$ whose closure is disjoint from $F_n$. But, you need to be a bit more careful, you need to select $a$ so that $ (\overline{U_1}\cup\overline{U_2}\cup\cdots\cup\overline{U_n}\cup\overline{U_{n+1}})^C\cap A$ is infinite. I'll leave this for you...