A Geometric Measure theoretic answer to your question could be the following.
Your example simply shows how things really go... There is nothing strange in it, nor in the "informal" notion of convergence you used: actually the trouble remains the same, even when all the underlying notions are well formalized.
The problem here is that the perimeter $\mathcal{P} (\cdot )$ is "only" a lower semicontinuous functional. This means: if a sequence of sets $E_n$ converges to another set $E$ in some sense (e.g. in the Hausdorff metric, which is a sort of uniform convergence for sets, or in some $L^p$ metric), then you have:
(*) $\displaystyle \mathcal{P} (E)\leq \liminf_{n\to \infty} \mathcal{P}(E_n)$;
moreover, in general the inequality is strict, even if the sequence at the RHside converges.
A less problematic example of this basic fact of GMT is the following. Let:
$E_n:=\{ (x,y)\in \mathbb{R}^2|\ x^2+y^2<1\text{ and } |y|\geq \frac{1}{n} |x|\}$,
so $E_n$ is the unit open circle with two symmetric slices (crossing the $x$ axis) removed and the slices become thinner and thinner as $n$ increases. Note that $E_n$ is a piecewise $C^1$ set with a finite number of corners, i.e. $5$, and this number does not increase with $n$ (on the contrary, in your example the number of corners becomes larger as $n\to \infty$).
Then $E_n$ converges to the unit open circle $D$ in the $L^1$ metric: in fact, the measure of the symmetric difference set $D\Delta E_n$ tends to $0$ as $n$ increases, i.e. $\lVert \chi_D -\chi_{E_n}\rVert_1\to 0$; on the other hand, one has:
$\displaystyle \mathcal{P} (E_n)=4+ 2\left( \pi -2\arctan \tfrac{1}{n}\right)$
(the summand $4$ appears because the boundary $\partial E_n$ contains four radii of lenght $=1$) therefore:
$\displaystyle \liminf_{n\to \infty} \mathcal{P}(E_n)=\lim_{n\to \infty} \mathcal{P} (E_n) =2\pi +4 >2\pi =\mathcal{P} (D)$.
For what is worth, $E_n$ converges to $D$ also in the strongest Hausdorff metric, because it is not hard to prove that the Haudorff distance:
$\displaystyle \text{dist}_H(E_n,D):=\inf \{ \epsilon >0| E_n\subseteq (1+\epsilon)D \text{ and } D\subseteq E_n+\epsilon D\}$
is given by:
$\displaystyle \text{dist}_H(E_n,D) =2-\frac{2}{\sqrt{1+\frac{1}{n^2}}}$,
and $\displaystyle \lim_{n\to \infty} \text{dist}_H(E_n,D) =0$.
If you choose $E_n$ as in your example, then you have the same situation: a sequence of sets which does converge to a triangle in some metrics (in particular, it converges in both $L^1$ and Hausdorff metric) and for which strict inequality holds in (*).
Here is an elegant proof, but it relies on a bit of real analysis and not just euclidean geometry. I assume you are concerned with only non-self-intersecting polygons since it is not clear what should be defined as the area of a self-intersecting polygon.
[Edit: I don't know how I misread the question to be about polygons inscribed in a given circle with the maximum area. The first proof solves that and the second proof solves the original problem.]
Cyclic Polygon Maximum Area
Let $D$ be the set of vectors in $\mathbb{R}^n$ that describe the angles subtended by the sides of a non-self-intersecting cyclic polygon (which is allowed to have sides of length $0$).
Then $D$ is clearly a closed bounded set and hence compact.
Let $f(v)$ be the area of the polygon $P$ that is described by the vector $v$.
Then $f$ is continuous on $D$.
Therefore $f$ attains a maximum on $D$.
Let $a \in D$ such that $f(a) = \max_{v \in D} f(v)$ and let R be the polygon described by $a$.
If $R$ is not regular:
$R$ has two sides $AB,BC$ of unequal length.
Move $B$ on the arc $AC$ to make $AB,BC$ of equal length.
Then $B$ is now further away from $AC$ and hence $Area(\triangle ABC)$ increases.
Thus $Area(R)$ increases.
Contradiction.
Therefore $R$ is regular.
Isoperimetric Polygon Maximum Area
Let $p > 0$ be the given perimeter.
Let $D$ be the set of points in $(\mathbb{R}^2)^n$ that describe the vertices of a directed polygon of perimeter $p$ (which is allowed to have sides of length $0$ and self intersections) such that one vertex is $(0,0)$.
Then $D$ is a closed (since it can be expressed as non-strict inequalities) bounded set and hence compact.
Let $f(v)$ be the signed area of the polygon $P$ that is described by the vector $v$.
Then $f$ is continuous on $D$.
Therefore $f$ attains a maximum on $D$.
Let $a \in D$ such that $f(a) = \max_{v \in D} f(v)$ and let R be the polygon described by $a$.
If $R$ has two sides $AB,BC$ such that $|AB| \ne |BC|$ or $Area(\triangle ABC) < 0$:
Move $B$ to preserve $|AB|+|BC|$ but make $|AB| = |BC|$ and $Area(\triangle ABC) \ge 0$.
The locus of $B$ that preserves $|AB|+|BC|$ is an ellipse with $AC$ as a diameter.
Thus $Area(\triangle ABC)$ increases and hence $Area(R)$ increases.
Contradiction.
Therefore $R$ has all sides of equal length and each internal angle being at most $180^\circ$.
If $R$ is not cyclic:
Let $X,Y,Z,W$ be four vertices of $R$ in order that do not lie on a circle.
These four vertices divide the sides of $R$ into four sections.
Move those four sections rigidly so that $R$ remains a polygon but $XYZW$ becomes cyclic.
Then $Area(R)$ increases by Bretschneider's formula.
Contradiction.
Therefore $R$ is cyclic.
Therefore $R$ is a regular polygon (possibly a star).
Thus $R$ has area $n·\tan(α/2)$ where $α$ is the internal angle between sides of $R$.
Thus $R$ is a regular polygon since it alone maximizes $α$.
Notes
The above proofs also show with essentially no change that the regular polygon is the only case with the maximum area. My second proof uses signed area since it was easier than requiring the polygon to be non-self-intersecting, but I'll leave my first proof the way it is.
Best Answer
Edit: I like the above figures because they're easy to generalize to many sides. But if it's unclear that they have the same area, here's another pair: the L and T tetrominoes.
You can imagine sliding the square on the right side up and down relative to the 1x3 bar on the left side; this operation preserves both area and perimeter. Explicitly, both tetrominoes have area 4 and perimeter 10. The L has 6 sides, and the T has 8 sides.