Show the following is a subspace and find its dimension If $V=K^{2009}$ where $K$ is a field. Show $W=\{(a,b,a,b,a,b,...)|a,b \in K \}$ is a subspace and find $dim_{K}W$.  
My Attempt;
For $W$ to be a subspace of $V$ two propeties must hold;
Closure by Additivity
Closure by Scalar Multiplication 
For the first assume two elements of ;
$v_1=(w,x,w,x,w,x,w,x,.....)$ for some $w,x \in K$
$v_2=(y,z,y,z,y,z,y,z,.....)$ for some $y,z \in K$
The sum of the two
$v_1+v_2=(w+y,x+z,w+y,x+z,w+y,....)$ where $w+y,x+z \in K$
So closure holds.
For the second assume an element of $W$ ;
$v_1=(w,x,w,x,w,x,w,x,.....)$ for some $w,x \in K$
and a scalar $\alpha \in K$
then scalar multiplication yields,
$\alpha v_1=(\alpha w,\alpha x,\alpha w,\alpha x,...)$ where $\alpha w , \alpha x \in K$
So closure by multiplication holds.
And therefore W is a subspace.  

Find $dim_{K}W$  

Now wouldn't the basis of $W$ consist of all the possible pairings of two unique members of $K$, wouldn't that be countably infinite? is this correct?

If you have two parameters that you are varying in your definition (your $a$ and $b$), this is a strong indication that your vector space has dimension $2$.  Now, we want to exhibit a basis of dimension $2$, so for this, think of the vectors:
$$( 0,1,0,1,\dots )$$
$$( 1,0,1,0,\dots )$$
Pay attention to the first component and second component to show linear independence, and then show that this spans the entire space.
