(a) Good to see you here, Paul.
(b) To address the missing link in Dan's solution (step 1):
Let $P$ be the set of all plane polygons with $n$ vertices, modulo congruence. We need a topological structure on $P$ such that the map $A:P\rightarrow\mathbb{R}$ that takes a polygon to its area is continuous, and such that the subset $P_\ell$ of $P$ consisting of polygons of a fixed perimeter $\ell$ is compact. Then, compactness of $P_\ell$ and continuity of $A$ together imply that $A$ attains a maximum value on $P_\ell$, and Dan's argument can get started.
I am sure there is a completely standard way to do this, but I don't know it, so I'll take it from scratch:
I feel that there is only one sensible choice for the topological structure on $P$, so we had better hope it has these properties.
Start with $(\mathbb{R}^2)^n$, viewed as the set of $n$-tuples of points in the plane $(p_1,\dots,p_n)$. (From here on, "coordinate" means a point in the plane, not a real number.) Each $n$-tuple characterizes some polygon uniquely: Start at $p_1$ and draw a line to $p_2$, etc., returning to $p_1$ at the end. However, the same polygon is characterized by different $n$-tuples: $(p_2,\dots,p_n,p_1)$ and $(p_n,p_{n-1},\dots,p_1)$ both give us the same polygon as $(p_1,\dots,p_n)$ for example. The amount of ambiguity is given precisely by the dihedral group $D_n$, acting on the coordinates, since we could start at any of the vertices and go in either direction to turn a polygon into an $n$-tuple of points. So we need to quotient $(\mathbb{R}^2)^n$ by the action of $D_n$ on the coordinates. (Here $D_n$ is being realized as the subgroup of $S_n$ generated by the long cycle $(1\dots n)$, and the permutation $(1n)(2(n-1))\dots$ that reverses the order of the coordinates.)
Thus the quotient $(\mathbb{R}^2)^n / D_n$ is the set of distinct polygons. To reduce to $P$ we need to mod out by congruence, thus to quotient by the group $E(2)$ of Euclidean motions of the plane. Since $D_n$ is permuting the coordinates and $E(2)$ is acting separately on each coordinate, their actions on $(\mathbb{R}^2)^n$ commute with each other, thus $P=(\mathbb{R}^2)^n/(D_n\times E(2))$.
The only sensible topology is the quotient topology.
Now, for sure, this topology does make the area map $A:P\rightarrow \mathbb{R}$ continuous. We can define an area map starting with $(\mathbb{R}^2)^n$ by drawing the polygon given by the $n$-tuple and then calculating its area; I don't feel the need to argue that this is continuous. But, this map factors through $P$ because the group $D_n\times E(2)$ acting on the $n$-tuples will not affect the area. The whole point of the quotient topology is that continuous maps from the original space that factor through the quotient are automatically continuous, so $A:P\rightarrow\mathbb{R}$ is continuous.
By the same logic, the perimeter map $p:P\rightarrow\mathbb{R}$ is also continuous, and the set $P_\ell$ of polygons of fixed perimeter $\ell$ is precisely $p^{-1}(\{\ell\})$. The set $\{\ell\}\subset \mathbb{R}$ is closed, so its preimage in $P$ under the continuous map $p$ is also closed; thus $P_\ell$ is a closed subset of $P$.
My plan is to exhibit a subset $A$ of $P$ that contains $P_\ell$ and is definitely compact. Then $P_\ell$, as a closed subset of the compact space $A$, will be compact.
Every polygon is congruent to one that has the origin as a vertex. No polygon whose perimeter is $\ell$ and has one vertex at the origin can have any of its other vertices outside of the disc of radius $\ell$ about the origin. Thus every polygon of radius $\ell$ is congruent to one that is represented by an $n$-tuple that looks like $(0,p_2,\dots,p_n)$ with $p_2,\dots,p_n$ all within the closed disc of radius $\ell$ about the origin. Let $A'$ be the subset of $(\mathbb{R}^2)^n$ consisting of $n$-tuples of this form: i.e. $A'$ is
$$\{0\}\times \overline{B}_\ell(0) \times\dots\times \overline{B}_\ell(0)$$
(where $\overline{B}_\ell(0)$ is the closed disc of radius $\ell$ about the origin). $A'\subset (\mathbb{R}^2)^n$ is clearly compact. But also, because every polygon of perimeter $\ell$ is represented by such an $n$-tuple, the image of $A'$ under the quotient map $(\mathbb{R}^2)^n\rightarrow (\mathbb{R}^2)^n/(D_n\times E(2))$ contains $P_\ell$. This image is the desired $A$. It is compact because $A'$ is compact, the quotient map is continuous (by construction of the quotient topology), and the continuous image of a compact set is compact.
This does it: now $P_\ell\subset A$ is a closed subset of a compact set, thus compact.
(c) Incidentally, I agree with your dimension count for $P_\ell$. In terms of the above discussion, $P$ is the quotient of $2n$-dimensional $(\mathbb{R}^2)^n$ by $3$-dimensional $D_n\times E(2)$; thus $P$ is $(2n-3)$-dimensional. ($E(2)$ is 3-dimensional because its quotient by its translation subgroup is the orthogonal group $O_2$, which is $1$-dimensional because it is homeomorphic to two copies of $S^1$, and the translations are $2$-dimensional because they are homeomorphic/isomorphic to $\mathbb{R}^2$. $D_n$ is finite, thus zero-dimensional, so doesn't affect the dimension.) Fixing the perimeter knocks off $1$ more dimension.
