$\pi_1(\mathbb{R}^2\setminus (\mathbb{Z}\times 0))$ I'm looking for the fundamental group of Y:= $\mathbb{R}^2\setminus (\mathbb{Z}\times 0) $. In fact,I'm not. I already know that it is the free group in countable generators, but I'm looking for a proof.
My aim is to consider the rectangles $A_n=\{(x,y)\in Y||x-n|\leq1/2\}$ because $Y=\bigcup A_n $ and I want to use this corollary of Seifert van Kampen: 

Let $X $ be a Hausdorff space, $A_i\cong \mathbb{S}^1$  some subspaces of $ >X$ indexed by some set $ I$ such that
  
  
*
  
*$X= \underset{i\in I}\bigcup A_i $.
  
*$A_i\cap A_j=\{x\} $ for every $i,j\in I $ and for some $x\in X $.
  
*A subset $C\subseteq X$ is open in $X$ (closed) if and only if $C \cap A_i $ is open (closed) in $ A_i$ for every  $i\in I$.
  
  
  Then the fundamental group of $X$ based on $x $ is the free group on $ I$

My idea is to retract each $A_n $ in to the circle centred in $n$ of radious $1/2 $ which is homeomorphic to the cirlce as required. Then I want to identify all the intersection points in the circles with a certain $x$ and take the quotient so that $A_n \cap A_n =\{x\}$ for every $m,n\in \mathbb{Z}$. By doing this I think the conditions in the box are satisfied but I'm not sure if the resulting space is homeomorphic to $Y$ although it seems like it is.
If this is not the way I dont know which is. I'm really having troubles with this one. Please help.
Thank you all in advance
Important Note:
I'm aware of the fact that someone has already asked this questionbefore, but I'm not comfortably with the answer they gave him because it uses grupoids and other concepts which I think are not necessary since this is a question from the firsts chapters of a basic book in algebraic topology (namely Massey).
 A: After a (big) while a friend of mine (Ángela Vargas) in my course of Algebraic Topology came up with a very beautiful and simple answer. She used the following generalization of Seifert Van Kampen:

If $X$ is a topological space, and $\{U_i\}_I\cup\{W\} $ is an open cover of open arcwise connected sets and $W $ is a simply connected set such that $U_i\cap U_j=W $ for every $i\neq j $, $W\subsetneq U_i\;\forall i\in I $ and $x_0\in W $ then $\pi(X,x_0) $ is the free product of $\pi(U_i,x_0)$.

For our space she took $ I=\mathbb{Z}$ and $ U_i$ to be the union of the circles I described in the question each without the lowest point except by the $i -$th circle. Each of the $U_i $ has $\mathbb{Z} $ as its fundamental group, and for any $i\neq j $ we have that $U_i\cap U_j $ is the union of infinitely many circles each without a point, which is simply connected. By the generalization above we have the result.
A: I think your union of circles is a deformation retract of $Y$, so the spaces are homotopy equivalent and have the same homotopy groups. 
A slightly better deformation retract of $Y$ may be taking circles of radius $1/4$ and one of their joint tangent lines. Then the set you need to quotient out becomes contractible.
