Alternatively, you can use the definition: a point is in the boundary of $N$ if and only if every open neighbourhood intersects with both $N$ and the complement $N^c$.
Proof: Let $x \in \partial \overline{N}$. We want to show that if $U$ is an open neighbourhood of $x$ then $U \cap N \neq \varnothing$ and $U \cap N^c \neq \varnothing$. Since $x \in \partial \overline{N}$ we have that $U \cap \overline{N} \neq \varnothing$ and $U \cap \overline{N}^c \neq \varnothing$ and since $U \cap \overline{N}^c \subset U \cap N^c$ we have $U \cap N^c \neq \varnothing$.
To show that $U \cap N \neq \varnothing$, we use that since $x \in \partial \overline{N}$, $U$ contains a point in $\overline{N}$ and a point in $\overline{N}^c$. Let $y \in U \cap \overline{N}$. Then in particular, $y \in U$, and since $U$ is open, $U$ is also a neighbourhood of $y$. Also, since $y \in \overline{N}$, every neighbourhood of $y$ contains a point $z \in N$. Hence $z \in U \cap N \neq \varnothing$ which proves the claim.