My proof of the theorem : finitely many disjoint discs can be inscribed in a unit square with total area approaching 1 This question has the answer here by Hagen von Eitzen. I consider the proof as unwantedly lengthy. I am looking to simplify the proof.
My proof:
Consider a convex shape $S$ of positive area $A$ inside the unit square. Let $a\le 1$ be the supremum of all subsets of the unit square that can be obtained as disjoint union of finitely many scaled and translated copies of $S$.
Partition the square into $n\times n$ smaller squares (see picture).
There are three types of such small squares: $e$ exterior squares (white in the picture), $i$ interior squares (light red in the image) and $b$ boudary squares (blue/purple). Of course $e+b+i=n^2$

$$\dfrac{i}{n^2} < A$$
$$\implies\dfrac{i}{n^2} + \dfrac{b}{n^2} < A + \dfrac{b}{n^2} \tag1$$
Picking a finite packing that covers $\ge a-\epsilon$, for some $\epsilon$, we can put a scaled-down copy of this packing into each of the $e$ "white" squares and, together with the original shape $S$, obtain a finite packing of the unit square that covers $e \dfrac{a- \epsilon}{n^2}  + A$
\begin{align}
\text{Area of finite disc packing}
&= e \dfrac{a- \epsilon}{n^2}  + A  \tag2
\end{align}

\begin{align}
a>\text{Area of finite disc packing}
&= e \dfrac{a- \epsilon}{n^2}  + A \tag {by 2}\\
&= \dfrac{e}{n^2} (a- \epsilon) + A \\
&= \left[   1- \left(   \dfrac{i}{n^2} + \dfrac{b}{n^2}  \right)  \right]   (a- \epsilon) + A\\
&\geq \left[   1- \left(   A + \dfrac{b}{n^2}  \right)  \right]   (a- \epsilon) + A \tag {by 1}\\
&=\left[   1 - A - \dfrac{b}{n^2}    \right]   (a- \epsilon) + A
\end{align}
$$\implies \left[   1 - A - \dfrac{b}{n^2}    \right]   (a- \epsilon) + A<a$$
As $n\to\infty$ and $\epsilon\to 0$ the LHS converges to $a+(1-a)A$. According to a limit theorem, this limit must be $\le a$. Thus we conclude $a=1$.

My question:
Is my proof correct and sufficient?

 A: No, your proof doesn't quite work.
The difference between your attempt vs Hagen's original proof is that they used the extra step $b \le 8(n-1)$ so their final expression before the limit step was $$a > \left(1-A-\frac{8}{n} \right) (a-\epsilon) + A.$$
At that point, it's valid to take the limits as $n \to \infty$ and $\epsilon \to 0$ and we're done.
In your proof, you've attempted to simplify the argument by skipping the $b \le 8(n-1)$ step. Your final expression before taking limits is
$$\left[   1 - A - \dfrac{b}{n^2}    \right]   (a- \epsilon) + A<a.$$
This time it isn't valid to take the limits you need, because your expression includes $b$ which depends on both $\epsilon$ and $n$ in some confusing way. What if $b$ turned out to grow extremely fast as you choose $n$ large and $\epsilon$ small?
This issue is why the other author went out of their way to bound $b$ so that they could create a final expression that just depends on $a, A, n$, and $\epsilon$ before taking limits. (It's fine to still have $a$ and $A$ in the expression because those don't depend on $n$ or $\epsilon$.)
