Whether I am correct or wrong I don't know. If there are any corrections, please let me know.
Let $p_n$ = product of all primes. (of course we can go still beyond as we know $p_n$ is infinite). Now consider $M = p_n + 1$ and $N = p_n - 1$. As we know that, by Euclid theorem, no prime of $p_n$ will divide $M$ as it leaves remainder $1$. So, $M$ is prime. At this same time, no prime of $p_n$ can divide $N$ as it leaves the remainder $1$. So, $N$ must be prime. Now, $M - N = (p_n +1 ) - (p_n -1) = 2$, i.e. $(M, N)$ is prime pair with difference $2$. So, if we extend primes still we can see the difference $2$. So twin primes are not finite.
(I will use the notation $p_n=$ the $n$th prime number, not $p_n=$ product of the first $n$ prime numbers as the OP is using.)
Your argument is incorrect, since neither $M=(p_1\cdots p_n)+1$ nor $N=(p_1\cdots p_n)-1$ have to be prime. All you know is that $M$ and $N$ are not divisible by the primes $p_1,\ldots,p_n$.
For some specific examples:
$$(2\cdot 3\cdot 5\cdot 7\cdot 11\cdot 13)+1=30031=59\cdot 509$$ $$(2\cdot 3\cdot 5\cdot 7)-1=209=11\cdot 19$$
As you can see, $209$ is not divisible by any of $2,3,5,7$, and $30031$ is not divisible by any of $2,3,5,7,11,13$, but this has not prevented them from being divisible by other primes, and hence composite.
See also the discussions here and here.