Q:

Spot the mistakes? Proof of the Twin Prime conjecture and Goldbach's theorem

The Twin Prime Conjecture
For any prime number $p_x$ larger than 3, there exists a number $n$ that is less than $p_x^2 -2$ and does not have a remainder of $\pm 1$ when divided by any prime number less than or equal to $p_x$.
Why is this the same as the Twin Prime Conjecture?
If $n$ exists as above, then $n \pm 1$ are prime numbers because; they are not divisible by any prime number less than or equal to $p_x$ and they are less than $p_{x+1}^2$ which is the smallest composite number whose prime factors are all greater than $p_x$. As there are infinitely many prime numbers, it follows that there are infinitely twin prime numbers.
Direct Proof 
If $n$ does exist as above. It follows that $n \equiv 0\mod 2$ and $n \equiv 0 \mod 3$ as all other values contradict the OP statement. It also follows that for any other prime number $p_i$, there are only $p_i - 2$ possible remainders that $n$ can have when divided by $p_i$, because the remainders "$\pm 1$" also contradict the OP statement.
The Chinese Remainder Theorem ensures the existence of a number $m$ that satisfies the system of conditions, $m \equiv a_i\mod p_i$, where $1 \leq i \leq x$ and $a_i$ are specified remainders.
The Chinese Remainder Theorem also gives a way of calculating the value of $m$.
By the Chinese Remainder Theorem, if $n$ exists as above, then 
$n \equiv \Sigma_{i=3}^x (a_i)(\frac{b_iP_x\#}{p_i}) \mod P_x\#$ where $P_x\# := \Pi_{i=1}^x p_i $ and $b_i$ is given by solving $\frac{b_iP_x\#}{p_i} \equiv 1 \mod p_i$.
Let $c_i = \frac{b_iP_x\#}{p_i}$, then $n \equiv \Sigma_{i=3}^x (a_i)(c_i) \mod P_x\#$
If $c_i \equiv 1 \mod p_i$ then $c_i \mod P_x\#$ can take any one of $\frac{P_x\#}{p_i}$, depending on which value of $b_i$ is chosen. 
Choose $b_i$ such that  $c_i \equiv 1 \mod P_x\#$
It follows that $n \equiv \Sigma_{i=3}^x (a_i) \mod P_x\#$.
Choose  $a_i = 2$, then $n < 2x <p_x^2 $
Q.E.D.
Goldbach's Conjecture
For any number $n$ such that $p_x^2 < 2n < p_{x+1}^2$ and $p_{x+1}^2 < P_x\#$, there exists another number $e < n - p_x$ such that $n \pm e$ are not divisible by any prime number less than or equal to $p_x$.
Direct Proof
For any prime number $p_i$ less than or equal to $p_x$, there are only certain remainders that $e$ can have when divided by $p_i$ to ensure that $n \pm e$ is not divisible by $p_i$. 
If $e$ exists, it follows from the reasoning in the last proof that $e \equiv \Sigma_{i=1}^x (a_i) \mod P_x\#$. 
For all $p_i$ that are not factors of $n$, choose $a_i = 0$. If $p_{x+1}^2 < P_x\#$ then there will be atleast one value of $i : a_i = 0$ because otherwise $p_{x+1}^2 < n$ which is a contradiction. 
$\Sigma_{i=1}^x p_i < n - p_x$ for all $n \geq 14$. Starting at $n=14$, when the LHS sum increases by $p_x + \theta$,  the RHS increases by an increment of $\theta p_x + \frac{1}{2}\theta^2$.
As $a_i < p_i$, and $\Sigma_{i=1}^x p_i < n - p_x$ it follows that an $e$ exists that satisfies all the conditions.
Q.E.D

A:

Your essential problem is that $b_i\cdot \frac{P_x\#}{p_i}\equiv 1\pmod{p_i}$ determines $b_i$ up to a multiple of $p_i$, so $b_i=\tilde b_i+kp_i$. Hence the possible values of $c_i:=b_i\cdot \frac{P_x\#}{p_i}$ are $\tilde b_i\cdot \frac{P_x\#}{p_i}+kP_x\#$ and so contrary to your argument does not take  different values modulo $P_x\#$.

