Part $1$.
If $\pi(n) \sim \frac{n}{\ln(n)}$ by the prime number theorem, can we treat $\frac{1}{\ln(n)} $ as the probability that a number less than $n$ is a prime number?
Say we have some operation $\Delta : n\Delta e = n-e,n+e $ so that $n\Delta_{+}e = 2n $, could we prove Goldbach's conjecture by showing that $ \frac{n}{\ln(n)\ln(2n)} \geq 1$ ?
Part $2$.
Suppose we know that in the interval $[1,p^2_x)$, the number of prime numbers is $ \geq\frac{2(p^2_{x}-1)}{p_{x}+1}$ and in the interval $[p^2_{x},p^2_{y})$, the number of prime numbers is $\geq \frac{2(p^2_{y}-p^2_{x})}{p_{x}+p_y}$.
Can we treat $\frac{2}{p_{x}+1}$ as the lowest probability that a natural number less than $p^2_{x}$ is prime, and $\frac{2}{p_{x}+p_y}$ as the lowest possibility that a natural number between $p^2_{x}$ and $p^2_{y}$ are prime, and assume these probabilities are independent of each other?
Could we then prove Goldbach's conjecture by showing that $\frac{2\times2\times (p^2_{x}-1)}{(p_{x}+1)\times(p_{x}+p_{y})} = \frac{4(p_{x}-1)}{p_{x}+p_{y}} \geq1$ when the following conditions are true:
a) $p^2_{x}\leq n<2n<p^2_{y}$
b) $p^2_{x}\leq n<p^2_{x+1}$ and $p^2_{x}<2n\leq 2(p^2_{x+1}-1) : \frac{4(p_{x}-1)}{p_{x}+2(p^2_{x+1}-1)} \geq 1$
c) $\exists p_{y}: 3p_{x}-4 \geq p_{y}> \sqrt{2(p^2_{x+1}-1)} \because \frac{4(p_{x}-1)}{p_{x}+p_{y}} \geq1\implies 3p_{x}-4 \geq p_{y}$
Here is a very simple reason why such approach cannot work, unless you combine it with very special properties of prime numbers.
Let
$$A := \{ 3 \lfloor \frac{n \ln n}{3} \rfloor | n \in \mathbb N \}$$
Note that the probability that a number is in $A$ is also $\frac{1}{\ln n}$, and every reasoning in your tentative proof also applies to this set $A$.
If your method is successful, then it also applies to this $A$, and can prove that any positive integer is the sum of two elements in $A$.
But all elements in $A$ are multiple of $3$....