Boundary of Polygon is Topological Boundary

Theorem
Let $P$ be a polygon embedded in $\R^2$.

Denote the boundary of $P$ as $\partial P$.

Let $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$ denote the interior and exterior of $\partial P$, when $\partial P$ is considered as a Jordan curve.

Then the topological boundary of $\operatorname{Int} \left({P}\right)$ is equal to $\partial P$, and the topological boundary of $\operatorname{Ext} \left({P}\right)$ is equal to $\partial P$.

Proof
Denote the topological boundary of $\operatorname{Int} \left({P}\right)$ as $\partial \operatorname{Int} \left({P}\right)$, and denote the topological boundary of $\operatorname{Ext} \left({P}\right)$ as $\partial \operatorname{Ext} \left({P}\right)$.

Topological Boundary is Subset of Boundary
From Boundary of Polygon is Jordan Curve, it follows that the boundary $\partial P$ is equal to the image of a Jordan curve.

From Jordan Polygon Theorem, it follows that $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$ are disjoint, open and path-connected.

From Set Open iff Disjoint from Boundary, it follows that $\operatorname{Int} \left({P}\right)$ and $\partial \operatorname{Int} \left({P}\right)$ are disjoint.

From Disjoint Open Sets remain Disjoint with one Closure, it follows that $\operatorname{Ext} \left({P}\right)$ and the closure of $\operatorname{Int} \left({P}\right)$ are disjoint.

As $\partial \operatorname{Int} \left({P}\right)$ is a subset of the closure of $\operatorname{Int} \left({P}\right)$, it follows that $\operatorname{Ext} \left({P}\right)$ and $\partial \operatorname{Int} \left({P}\right)$ are disjoint.

As $\R^2 = \operatorname{Int} \left({P}\right) \cup \operatorname{Ext} \left({P}\right) \cup \partial P$ by the Jordan Polygon Theorem, it follows that $\partial \operatorname{Int} \left({P}\right) \subseteq \partial P$.

Similarly, it follows that $\partial \operatorname{Ext} \left({P}\right) \subseteq \partial P$.

Boundary is Subset of Topological Boundary
Let $p \in \partial P$ such that $p$ is not a vertex, and let $S$ be the Definition:Side of $P$ that $p$ is a part of.

Denote the $j$th side of $P$ as $S_j$, and let $n \in \N$ be the total number of sides.

Let $\displaystyle \delta = d \left({S, \bigcup_{ j = 1, \ldots, n: S_j \ne S } S_j }\right)$ be the Euclidean distance between $S$ and all other sides of $P$.

From Distance between Closed Sets in Euclidean Space, it follows that $\delta > 0$.

Let $\epsilon \in \left({0\,.\,.\,\delta}\right)$, and denote the open ball of $p$ with radius $\epsilon$ as $B_\epsilon \left({p}\right)$.

Choose $x_1 \in B_\epsilon \left({p}\right)$, and put $\mathbf v = p - x_1$.

Let $\mathcal L_1 = \left\{ {x_1 + s \mathbf v: s \in \R_{\ge 0} }\right\}$ be a ray with start point $x_1$.

Then $\mathcal L_1$ and $S$ has one crossing at $p$.

Put $x_2 = x_1 + 2 \mathbf v$, and put $\mathcal L_2 = \left\{ {x_2 + s \mathbf v: s \in \R_{\ge 0} }\right\}$, so $\mathcal L_1 \cap \mathcal L_2 = \mathcal L_2$.

Then $\mathcal L_2$ and $S$ do not cross.

As $x_2 \in B_\epsilon \left({p}\right)$ with $\epsilon < \delta$, it follows from the definition of $\delta$ that if $\mathcal L_1$ and some side $S'$ has a crossing, then $\mathcal L_2$ and $S'$ also has a crossing.

If $N \left({x_i}\right)$ denotes the number of crossings between $\mathcal L_i$ and $\partial P$, it follows that $N \left({x_1}\right) + 1 = N \left({x_2}\right)$.

Then $\operatorname{par} \left({x_1}\right) \ne \operatorname{par} \left({x_2}\right)$, where $\operatorname{par} \left({x_i}\right)$ denotes the parity of $x_i$.

From Jordan Polygon Interior and Exterior Criterion, it follows that one of the points $x_1, x_2$ belongs to $\operatorname{Int} \left({P}\right)$, and the other point belongs to $\operatorname{Ext} \left({P}\right)$.

As $\epsilon$ was arbitrary small, it follows that $p$ is a limit point of both $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$.

By definition of closure, it follows that $p$ lies in the closure of $\operatorname{Int} \left({P}\right)$ and $\operatorname{Ext} \left({P}\right)$.

Then $p \in \partial \operatorname{Int} \left({P}\right)$ and $p \in \partial \operatorname{Ext} \left({P}\right)$, as the Jordan Polygon Theorem shows that $\partial P$ and $\operatorname{Int} \left({P}\right)$, $\operatorname{Ext} \left({P}\right)$ are disjoint.

Now, suppose that $p$ is a vertex of $S$.

Then we can find a sequence $\left \langle {p_k} \right \rangle$ of points that lies on the adjacent sides of $p$ such that the sequence converges to $p$.

As none of the point in $\left \langle {x_k} \right \rangle$ are vertices, all $x_k$ lie in $\partial \operatorname{Int} \left({P}\right)$ and $\partial \operatorname{Ext} \left({P}\right)$.

As Boundary of Set is Closed, it follows that $p \in \partial \operatorname{Int} \left({P}\right)$, and $p \in \partial \operatorname{Ext} \left({P}\right)$.

Hence, $\partial P \subseteq \partial \operatorname{Int} \left({P}\right)$, and $\partial P \subseteq \partial \operatorname{Ext} \left({P}\right)$.

The result now follows from Equality of Sets.