TITLE: Is this a valid argument for a prime always found between $p_n$ and $2p_n$ when $p_n > 25$
QUESTION [6 upvotes]: I recently came up with an argument for a subset of Bertrand's Postulate that I have not seen anywhere else.  Perhaps, it is not valid.  It is very short.  Please let me know if I have made a mistake.
(1) Let $p_n > 50$ be the $n$th prime.
(2) Let $p_n -c$ be the highest integer that is divisible by $6$ and less than $p_n$ so that $c$ is either $1, 3,$ or $5$
(3)  Let $S$ be the set of all integers $x$ such that $p_n < x < 2p_n$ and $\gcd(x,p_n-c)=1$
(4) Let $U \subseteq S$ be the set of integers $x$ such that $x \in S$ but $x$ is not a prime.  Let $u$ be the number of elements in $U$.
(5) All such $x$ in $U$ must have a least prime factor $q$ such that $5 \le q < \sqrt{2p_n}$.
(6) In all cases, $5 < \frac{x}{q} < p_n$ so that $p_n < p_n - c + \frac{x}{q} < 2p_n$ and $\gcd(p_n-c+\frac{x}{q},p_n-c)=1$  So, to be clear, for each of the $u$ non-primes, there are $u$ integers of the form $\frac{x}{q}$ where $p_n-c + \frac{x}{q}$ is between $p_n$ and $2p_n$ and $(p_n-c + \frac{x}{q})$ is relatively prime to $p_n-c$. 
(7) Since $p_n < p_n -c + p_n < 2p_n$, we have the interesting situation where there are $u$ elements of $S$ that are non-primes but at least $u+1$ elements in $S$.  So, one number between $p_n$ and $2p_n$ must be prime.   
(8) To be clear, the $u+1$ consists of the $u$ integers of the form $(p_n-c + \frac{x}{q})$ and the $1$ integer of the form $(p_n-c + p_n)$.  All $u+1$ are relatively prime $p_n-c$ and all $u+1$ are between $p_n$ and $2p_n$ but only $u$ are non-primes.

Edit: I made a change in step #1 in response to a comment by coffeemath.
Thanks, Coffeemath!
In order for $5 < \frac{x}{q} < p_n$, it is necessary that $p_n > 50$ 
With this correction, $\frac{p_n}{\sqrt{2p_n}} < \frac{x}{q} < p_n$ and:
$\frac{\sqrt{p_n}}{\sqrt{2}} > \frac{5\sqrt{2}}{\sqrt{2}} = 5$ 

Edit 2:  Added step (8) in attempt to make the $u+1$ and the $u$ count much clearer.
I made this change in response to questions asked in the comments.

Edit 3: I am not able to get past the observation made by coffeemath.   I am accepting  coffeemath's answer.

REPLY [2 votes]: Suppose $p_n=53>50.$ Then $c=5$ and $p_n-c=48=2^4\cdot 3.$ So $S$ consists of $$55,59,61,65,67,71,73,77,79,83,85,89,91,95,97,101,103.$$
The elements of $U$ are the non-primes of this list, namely
$$55,65,77,85,91,95.$$
Dividing each by its least prime factor gives respectively $11,13,11,17,13,19.$ These are the values of $x/q$ for $x \in U.$ Note they are not all distinct; two repeats. When we then calculate $p_n-c+x/q$ we get the list
$$59,61,59,65,61,67.$$
Again we have nondistinct values, with $59,61$ there twice each. [also the $65$ isn't prime, don't know if that's the problem...] The repeats seem to mean a double count has occurred.