What is wrong with this reasoning regarding twin primes? There are infinitely many primes $p$ of the form $6m - 1$. $p + 2$ is of the form $6m + 1$. Since there is a one-to-one correspondence between $p$ and $p + 2$ there are infinitely many $p + 2$ of the form $6m + 1$. There are infinitely many primes $q = p + 2$ of the form $6m + 1$. Thus, there are infinitely many corresponding pairs of primes $\{p, q\} = \{p, p + 2\}$. Thus, there are infinitely many twin primes.
 A: Two infinite subsets of $\Bbb N$ need not intersect.
A: (For each prime $p$ of the form $6m-1$) there is only one choice for $q = p+2$, and it might not even be prime. It will satisfy $q = 6m+1$, though.
There are infinitely many primes $q$ of the form $q = 6m+1$, but there is no (known provable) reason to expect many of those $m$ to be the values for which $6m-1$ is prime.
A: Here is a similar scenario:
There are infinitely many numbers of the form $6m-1$ which are divisible by $5$. And there are also infinitely many numbers of the form $6m+1$ which are divisible by $5$. But there are not infinitely many pairs $(6m-1,6m+1)$ which are both divisible by $5$. (In fact there are no such pairs, because $6m-1$ and $6m+1$ differ by $2$, so they can't have a common factor of $5$. But that isn't as important right now.)
Here is another similar one:
There are infinitely many numbers of the form $6m-1$ which are divisible by $7$. And there are also infinitely many numbers of the form $6m+1$ which are divisible by $7$. But there are not infinitely many pairs $(6m-1,6m+1)$ which are both divisible by $7$.
In general, for any property $P$, even if it is true that there are infinitely many $6m-1$ with property $P$, and infinitely many $6m+1$ with property $P$, it doesn't necessarily follow that there are infinitely many pairs $(6m-1,6m+1)$ that both have property $P$.
There could be "bad luck", if the values of $6m+1$ that have property $P$ are "interleaved" with the values of $6m-1$ that have property $P$, instead of "lining up".
Nobody knows if the property $P$ of primeness exhibits "interleaving" or "lining up" behavior. The Twin Prime Conjecture is that it "lines up" for infinitely many pairs $(6m-1,6m+1)$. But for now that is still open.
A: Let $P$ but the set of all primes.  Let $S$ be the set of all numbers that are $2$ less than a prime.
There are an infinite numbers of $p+2$ where $p \in P$.  And there are an infinite numbers of $p+2$ where $p \in S$.  But the $p$s that are in $P$ may be completely different than the $p$s that are in $S$.  
Yes for ever $p \in P\cap S$ we will have $p, p+2$ are a prime pair.  But we have utterly no reason to believe $P\cap S$ is infinite.
Replace "prime" in your argument with "is a multiple of 4":
There are infinitely many $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$ p of the form $\color{blue}5m−1$. p+2 is of the form $\color{blue}5m+1$. Since there is a 1-1 correspondence between p and p+2 there are infinitely many p+2 of the form $\color{blue}5m+1$. There are infinitely many  $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$ q=p+2 of the form $\color{blue}5m+1$. Thus, there are infinitely many corresponding pairs of  $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$ {p, q} = {p, p+2}. Thus, there are infinitely many twin  $\color{gray}{\text{primes}}$ $\color{blue}{\text{multiples of 4}}$.
But we no darn well there are $0$, not infinite, number of multiples of 4$ pairs.
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Actually here is a paraphrase of you argument.
Let $J =\{6m - 1|m \in \mathbb N\}$, $K=\{6m+2|m\in \mathbb N\}$ and let $j:J\to K$ via $j(n) = n+2$ by a bijection.
Let $P_1 = \{p\in J|p \text{ is prime}\}$ and $Q_1 = j(P_1) = \{j(p)|p \in P_1\}$.  $P_1$ is infinite and in 1-1 corespondence with $Q_1$.
Let $P_2 = \{p \in J|j(p) = p + 2 \text{ is prime}\}$ and $Q_2 = j(P_2) = \{q \in K| q \text{ is prime}\}$.  $Q_2$ is infinite and in 1-1 corespondence with $P_2$.
Let $P_3=\{p \in P_1|j(p) + 2 \in Q_2\}$ and $Q_3=\{q=j(p)=p+2|q \in Q_2; p \in P_1\}= j(P_3)$.  $P_3$ and $Q_3$ are in $1-1$ corespondence and for any $p \in P_3$ the pair $\{p, j(p)\}$ are a pair of twin pairs
Your argument is as follows:  $P_3 \subset P_1$ which is infinte;  $Q_3 \subset Q_2$ which is infinite.  $P_3$ and $Q_3$ are in 1-1 corespondence.  And they are both subsets of  infinite sets.  Hence they are infinite.
Which simply doesn't work.  In calling everything $p$ and $p+2$ and noting different sets wer infinite and in 1-1 corespondence of different sets, you simple got confused and wrongly assumed the wrong set coresponded to the wrong sets.
A: All you've done is assume that $p+2$ is prime.  So if you have $p := 6k - 1$, your assumption fails for $k = 4$, although $6k - 1 = 23 \in \mathbb{P}$ you should immediately see that $6k + 1 = 25 \not \in \mathbb{P}$.  Those are two different sets you have that are formed independently of each other.
I might refer you to these:
Prime Gaps in Residue Classes
...De Polignac Sequence
Asymptotic Expression...
so that you understand that we define these two sets similarly.  They have the same asymptotic, but the underlying structures are completely different. 
