After dissecting the mechanics of locating valid twin primes, I was able to establish a Proof through contradiction. I start by creating a table to easily display the potential list of twin primes. Using an elimination matrix scheme, I systematically remove twin prime candidates from the list if either half of the pair are multiples of an already known 'Prime Number'. Multiples of prime numbers, primes squared and primes multiplied by other primes are not prime numbers themselves (examples 5*5=25; 5*7=35; 5*11=55; 7*7=49; 7*11=77; and so on). It's an easy approach with repeatable patterns for each prime number. It quickly becomes obvious that these elimination patterns are repeating for all non-prime removals. All these elimination patterns are of the form remove-skip(n)-remove-skip(m)...repeated to infinity. Note that n+m+2 is the prime number. The first non-prime removal for any prime is in essense that prime^2 (prime squared). A prime number squared will always fall into the sixth column ( the column starting with 7)! Further, two adjacent patterns will slightly overlap if those two primes form a twin prime pair. I then proceed to make the 'silly' assumption that there will be no potential twin prime candidates in the initial skip(n) plus skip(m) regions for the two overlapped twin primes (entire initial pattern for a given twin prime pair) in this elimination matrix. If we assume that 11 & 13 are the last twin primes possible, we would have to make the assumption that there are no twin primes candidates in the elimination overlapped pattern regions for either prime 11 or 13 combined at minimum. The contradiction arises because we can show that there is always at least one twin prime pair in this combined/overlapped region. As long as there are infinitely many primes there will be infinitely many twin primes. Euclid proved there are infinitely many primes with his proof. I have simply extended his proof into my own.