Just to be clear: I don't think that this proof is valid; both it and I are far to simple to have proven the twin prime conjecture. I am only unable to find any mistakes in it.
Let $a$ be any prime number for which $a\le n$. All numbers $s$ for $n<s\le n^2$ where $s,n\in\mathbb{N}$ must be prime if they are coprime to all values of $a$.
This can be proven by contradiction: let $q$ be the smallest composite number which cannot be written as the product of possible values of $a$. Hence, $q$ is the square of the smallest prime that cannot be a value of $a$. As this number must, by definition, be greater than $n$, therefore $q>n^2$ and is thus outside of the domain.
From this, it can be concluded that any twin primes with midpoint $t$ can be found by verifying that $t\pm1$ is coprime to all values of $a$ where $n<t<n^2$. To verify whether a number $t$ is the average of a pair of twin primes, one must simply show that $t+1$ and $t-1$ are both coprime to all values of a.
For any given value of $a$ and $t$, if $t+1$ is divisible by $a$, $t+1\pm ma$ must also be divisible by $a$. If $t-1$ is divisible by $a$, $t-1\pm ma$ must also be divisible by $a$.
It should be noted that for any given value of $a$ and $t$, other than $a=2$, at most one of $t-1$ and $t+1$ can be divisible by $a$. Hence, the proportion of values of $t$ for which both $t+1$ and $t-1$ are coprime to a given value of $a$ other than $2$ is given by $\frac{a-2}{a}$. The requirement that $t\pm1$ be coprime to $2$ can simply be translated into the requirement that $t$ is even.
The domain $n<s\le n^2$ contains exactly $n^2-n-1$ consecutive integers. Of these, at minimum $\frac{n^2-n-2}{2}-1$ or $\frac{n^2-n-4}{2}$ must be even. Hence, more than $\frac{n^2-n-2}{2}-2=\frac{n^2-n-6}{2}$ must be even. Such minimums can be established for all possible values of a using the proportion $\frac{a-2}{a}$. Using this, one can establish that at minimum $\left(\left(n^2-n-2\right)\times\frac{3-2}{3}\right)-2=\frac{n^2-n-2}{3}-2=\frac{n^2-n-8}{3}$ numbers within the domain are such that both adjacent numbers are coprime to $3$. Furthermore, these expressions can be combined as $\left(\left(n^2-n-2\right)\times\frac{1}{2}\times\frac{3-2}{3}\right)-2-2=\frac{n^2-n-22}{6}$ to find the minimum amount of numbers adjacent only to numbers coprime to both $2$ and $3$. This process can be continued for all possible values of $a$, ending in the function $$N\left(n\right)=\left(\frac{n^2-n-2}{2}\times\prod_{k=3}^{n}\left(\frac{k-2}{k}\right)\right)-2A|k\in\mathbb{P}$$ This function provides the minimum number of twin primes within the domain $n<s<n^2$. A is defined as the number of possible values of $a$. This function can then be simplified: $$N_1\left(n\right)=\left(\frac{n^2-n-2}{6}\times\prod_{i=2}^{\frac{n}{2}}\left(\frac{2\left(i-1\right)}{2i}\right)\right)-2A$$
By removing the limitation that $i$ must be prime, the amount of terms that are being multiplied increases. The increased number of terms being multiplied in the product necessitates that $N_1(p)<N(p)$ as every term $j$ of the product fulfills the inequality $0<j<1$. The function $N_1(p)$ can then be simplified to $$N_1\left(n\right)=\frac{n^2-n-2}{6}\times\frac{\left(\frac{n}{2}-1\right)!}{\frac{n}{2}!}-2A$$ and hence $$N_1\left(p\right)=\frac{n^2-n-2}{2}\times\frac{2}{n}-2A=\frac{n^2-n-2}{n}-2A$$ At this point an overestimate for $A$ in terms of $n$ is required to simplify further. This can very crudely be done using $A<\frac{n}{4}$; this inequality is true for all $n>200$. Hence, $N_1\left(n\right)$ can be approximated by: $$N_2\left(n\right)=n-\frac{1}{n}-2\frac{n}{4}=\frac{n}{2}-\frac{1}{n}$$ It can clearly be seen that the function $N_2(n)$ diverges positively when $n\rightarrow\infty$. Due to the previously obtained inequality $N_2(n)<N_1(n)<N(n)$, $N(n)$ must also diverge positively for $n\rightarrow\infty$. Hence, there must exist infinitely many values $t$, such that $t$ denotes a prime number.
Note: This proof could be expanded to prove that for any even number, an infinate number of primes are seperated by this number. However, it is presented in its current form to make the process of finding any mistakes easier.
Where you write “Furthermore, these expressions can be combined”, you seem to be assuming some form of independence such that if $\frac12$ the numbers are not eliminated due to factors of $2$ and $\frac13$ are not eliminated due to factors of $3$, then at least $\frac16$ are not eliminated by either. I don't see a reason for this independence, and you don't seem to provide one. The numbers that are eliminated due to factors of $3$ could mostly be different from the ones eliminated due to factors of $2$; in this case all numbers could be eliminated in only two steps.
This sort of independence does hold “in the long run”. By virtue of the Chinese remainder theorem, if you take a stretch of numbers whose length is a product of primes, you'll get each combination of residues with respect to these primes exactly once, so in this stretch the proportion of numbers not eliminated by any of these primes is as you calculated it – but the product of all primes below $n$ is far greater than $n^2$, so this doesn't support your argument.