Is the following solution to the isoperimetric problem correct? I came up with a solution over here: http://keplerlounge.com/ 
but I am not entirely sure that it's free from error as much as 
I have checked it many times. If people can offer constructive
feedback I would really appreciate that.
Update1: @hardmath has brought to my attention that it's expected that 
I reproduce the proof on the forum. If that's the only issue that 
remains, here's the proof: 
1) A minimal condition on any closed curve C that satisfies the isoperimetric inequality
is that it must be convex.  A necessary property of any closed curve $C$ in order that for $A(C) = \underset{C}{\text{max}} A(C) $ is that the closed curve be convex. In consequence, this may be
calculated using any interior-point with the following integral:
{expression} . This may be deduced using Steiner reflection.
Without loss of generality, O = (0,0) is chosen to be the origin and for the remainder of
the proof I'll assume that we're dealing with a closed convex curve with finite perimeter
that will be denoted by $\displaystyle \gamma \displaystyle $.
2) Without loss of generality, any such curve $\displaystyle \gamma \displaystyle $ may be parametrized by time and inflated.
a) By inflation I mean that all proportions are kept equal while the area enclosed by the curve
increases. That is to say:
i) $ \delta t > 0 \iff r(\theta, t+\delta t) > r(\theta,t) $
ii) $f(t)$ is monotonically increasing and $r(\theta,t) = f(t)r(\theta)$
iii) $f(0) = 0$
iv) $\displaystyle \lim_{t \to +\infty}\displaystyle $ $\displaystyle f(t)=\infty\displaystyle $
b) The perimeter and area enclosed by the curve may be approximated by the lower-bounds:
$P_{m}(t)=2 \pi r_{m}(t) \leq P_{\gamma}(t)$
$A_{m}(t)=\pi r_{m}^2(t) \leq A_{\gamma}(t)$
where $r_{m}(t) = \underset{\theta}{\text{min}} r(\theta,t) $
c) Clearly, while the curve $\displaystyle \gamma \displaystyle $ is being inflated $\displaystyle P_{\gamma}\displaystyle $and $\displaystyle A_{\gamma}\displaystyle $ are both monotonically increasing functions, and we have the following boundary conditions on any curve $\displaystyle \gamma \displaystyle $
that is being inflated:
i) $\displaystyle P_{\gamma}(t=0)=0\displaystyle $
ii) $\displaystyle A_{\gamma}(t=0)=0\displaystyle $
iii) $\displaystyle \lim_{t \to +\infty}\displaystyle $ $\displaystyle P_{\gamma}(t)=\infty\displaystyle $
iv) $\displaystyle \lim_{t \to +\infty}\displaystyle$ $\displaystyle A_{\gamma}(t)=\infty\displaystyle $
3) Due to the finite perimeter property of $\displaystyle \gamma \displaystyle $ we have $\displaystyle \frac{dr(\theta,t)}{dt} < \infty\displaystyle $
If we define $\displaystyle f(\theta,t) =\frac{dr(\theta,t)}{dt} \displaystyle $, it follows that for any finite time interval $I = [0,k]$
there exists an M such that $\displaystyle \underset{\theta,t}{\text{max}} f(\theta,t) \leq M \displaystyle $ on that interval.
4) Now, in order to maximize the area enclosed by $\gamma$ we need to find $r(\theta,t)$ such that the distance travelled by points on the perimeter of $\gamma$ is maximized over the time-interval $I = [0,k]$. 
This is a simple optimisation problem:
max: $G(k) = r(\theta)(f(k)-f(0))$
constraint: $r(\theta)f'(t) \leq M$
The answer is straightforward. It would suffice to choose $f'(t) = M/r(\theta)$
Then we have $f(t) = Mt/r(\theta) + Cst$ and since $f(0) = 0$ this simplifies to the unique solution: $f(t)r(\theta) = Mt$.
5) From this point onwards, we may easily deduce the isoperimetric inequality:
$r(\theta,t) = f(t)r(\theta)=Mt$ is differentiable for all $t$ in $I=[0,k]$ so the perimeter and area may be calculated using: 
$\displaystyle P_{\gamma}(t) = \int_0^{2\pi}\sqrt{r(\theta,t)^2 - (\frac{dr(\theta,t)}{d\theta})^2} d\theta \displaystyle$
and $\displaystyle A_{\gamma}(t) = \frac{1}{2}\int_0^{2\pi} r(\theta,t)^2 \,d\theta \displaystyle $
This simplifies to $P_{\gamma}(t) = 2\pi Mt$ and $A_{\gamma}(t) = \pi (Mt)^2$
Furthermore, we must note that on the interval $I=[0,k]$ all points on $\gamma$ are
equidistant with respect to the origin so $\gamma$ is a circle. 
As for the inequality: 
In general, $\displaystyle A_{\gamma}(t) \leq \pi\frac{P_{\gamma}(t)}{4\pi^2}^2 \displaystyle $ so we obtain, $\displaystyle 4\pi A_{\gamma} \leq P_{\gamma}^2 \displaystyle $
Update2: The comments of @hardmath have been taken into account when editing this demonstration on 07/09/2014.
Update3: Today(29/09/2014) a professor at my university clearly explained that there were several serious holes in my proof. After his explanation I had to agree, and I must apologize to the people such as @hardmath who have patiently tried to explain my errors. I think I was too caught up in my own enthusiasm. 
I'm not sure whether this question should be removed. But, I think that the discussion on this page and the way it ended could serve other members of this community. 
 A: $0$. We are missing a proposition, the isoperimetric inequality, to be proven.  I want to reach the meat of the argument, but skipping preliminaries is an invitation to misunderstanding.  Consider how best to state the proposition to be proven.
$1$. The first step mentions a (simple?) closed curve $C$ (in the plane), and essays a reduction to the case of convexity and a parameterization of $C$ by polar angle $\theta$:
$$ r(\theta) \gt 0 \; \text{ for } \; 0 \le \theta \le 2\pi $$
where $r(0) = r(2\pi)$ is periodic.  There is also introduced an alternative symbol $\gamma$ for the closed curve $C$, perhaps in anticipation of the "inflation" technique which seems to involve a second parameter, $\gamma_t$ defined by function $r(\theta,t)$.
There are several objections I would make here.  No detailed argument is given about reducing to a convex region, and the statement of this is flawed: "A minimal condition on any closed curve $C$ that satisfies the isoperimetric inequality is that it must be convex."  Surely it is the bounded interior of curve $C$ that is to be assumed convex, and rather than saying it's a "minimal condition" of curves that satisfy the inequality (non-convex regions with finite perimeter also satisfy it), an argument similar to Steiner's reflection is wanted.  If we must preserve $\mathcal{C}^1$ smoothness of the boundary (depends on statement of proposition), some care is needed in invoking that argument.
My view is that these flaws are "fixable", and we can proceed tentatively to consider the next step.
$2$. The notion of "inflation" of the curve $C = \gamma$ is introduced with the partial explanation "all proportions are kept equal while the area enclosed by the curve increases."  This doesn't seem to be a complete specification of the family of curves $\gamma_t$ generated by the new "time" parameter, but if it is, it would seem to require that $r(\theta,t) = t r(\theta)$.
At any rate such an isotropic expansion seems to meet all the criteria for "inflation" that are outlined:
(a) $\delta t > 0 \iff r(\theta, t+\delta t) > r(\theta,t)$
(b) perimeter $P_{\gamma}(t=0)=0$ and $\lim_{t \to +\infty} P_{\gamma}(t)=\infty $
(c) area $A_{\gamma}(t=0)=0$ and $\lim_{t \to +\infty} A_{\gamma}(t)=\infty $
This seems to me an unfixable obstacle, unless some criterion for "inflation" has been overlooked (by me) or left unstated (by you).  A simple dilation $t$ of the curve $C$ will increase the perimeter by a factor $t$ and the area by a factor $t^2$.  This only conserves the satisfaction or dissatisfaction of the isoperimetric inequality as $t$ varies.  It cannot be the basis for asserting a proof of it.
$3$. Here it is asserted that $r(\theta,t)$ has a uniformly bounded (partial?) derivative with respect to time $t$ on compact intervals $[0,k]$.  This would certainly be so for the case of dilation $r(\theta,t) = t r(\theta)$ discussed above, so you can arrange for it to be so.  However it does not seem to follow from the little that was assumed about "inflation" (and no argument is advanced that justifies the claim).
If this is really needed for the argument, I'm willing to grant it is likely "fixable" by providing more specific definition of $r(\theta,t)$.
$4$. At "this point onwards" you declare success, "we may easily deduce the isoperimetric inequality," and move on.  But the immediate "iff" statement does not seem to represent a statement of the isoperimetric inequality.  The left hand side, $||r(\theta,t)|| \le Mt$ (why are we taking a norm, rather than simply absolute value, or even leave $r$ as is since it is positive?), follows from a Mean Value Thm. application of step $3$. above, but the corresponding bound on $A_\gamma$ does not give any comparison to $P_\gamma$.
In the next line we have a blatant switcheroo, where instead of the inequality $||r(\theta,t)|| \le Mt$ we suddenly have an equality $||r(\theta,t)|| = Mt$, and "$\gamma$ is a circle".
To summarize: I don't see what technique is being developed that will establish even the "easy" part of the isoperimetric inequality (the inequality itself), much less the more difficult part (that equality is attained only by a circle).
Added in response to Question Edits:
In the new step 2 a function $f(t)$ is introduced as defining the "inflation" process, i.e. $r(\theta,t) = f(t)r(\theta)$, with the apparent meaning that factor $f(t)$ will depend on time $t$ but not on angle $\theta$.
However in new step 3, "we define $f(\theta,t) = \frac{dr(\theta,t)}{dt}$".  Not only does this open the door to a function that appears to depend on $\theta$ as well as $t$, it confuses the issue of partial vs. ordinary derivatives.
This confusion becomes acute in new step 4, where "it would suffice to choose $f'(t) = M/r(\theta)$".  This choice is not possible unless $f(t)$ is allowed to be a function of $\theta$ as well as $t$.
Moreover to the extent that $f(t)$ varies with $\theta$, the "inflation" cannot provide that "all proportions [of the curve] are kept equal".  The inconsistency is manifest in new steps 4,5 where it is claimed:
$$ r(\theta,t) = f(t)r(\theta) = Mt $$
That is, the curve no longer depends on $\theta$ but only on $t$, i.e. the curve was a circle all along.  The perimeter and area of the curve are then "simplified" to the usual formulas for a circle, giving the appearance that the isoperimetric inequality has been proven.  However the argument is fallacious in suppressing the dependence of the family of curves on $\theta$.
