A new proof of Dirichlet's theorem concerning prime numbers in arithmetic progressions

It is known that there are infinitely-many prime numbers which take the form of a polynomial of degree one with integer coefficients, this is Dirichlet's theorem. We use an elementary sieving argument together with bounds on the prime number counting function to provide a new proof of Dirichlet's theorem. We show that if $a\in \mathbb{N},k\in \mathbb{N},a_{k}=(a,a+1,...,a+k-1)$ and $A=\left\{ p_{1},p_{2},...,p_{n}\right\}$, a finite set of primes. Then the number of components of $a_{k}$ that are divisible by some prime in $A$ is less than or equal to $$\sum\limits_{\substack{ d|P(A)\\ d>1}}(-1)^{\omega \left( d\right) +1}\left\lfloor \frac{k}{d}\right\rfloor +2n$$ where $\omega \left( d\right)$ is the number of distinct prime divisors of $d$ and $P(A)=\prod_{p\in A}p$. We claim that the $+2n$ in the bound can be replaced with $n$, the \texttt{best possible bound}. However, we did not demonstrate our claim in this paper since the $+2n$(bound) is enough for the new proof of Dirichlet's theorem. This result effectively means that given $[1,x], x\in\mathbb{R}$; if the primes in $A$ divide $h$ integers in $[1,x]$ then for every $g>0$, they will divide at most $h+2|A|$ integers in $[1+g,x+g].$