Why is $H^n(\mathbb{R}P^2;\mathbb{Z}) \otimes H^n(S^1;\mathbb{Z}) \not =0$ in degree $n = 1,2$? 
Compute $$H^*(\mathbb{R}P^2\times \mathbb{R}P^2, \mathbb{Z}).$$
Hint: Consider the Mayer-Vietoris sequence of suitable open sets of the form $\mathbb{R}P^2\times U$ and $\mathbb{R}P^2\times V$ where
$U$ is a small disk in $\mathbb{R}P^2$.

So i was trying to understand the proposed solution to this exercise. But building the Mayer-Vietoris sequence goes wrong since i somehow end up getting $$H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z}) = 0$$
which is apparently wrong.

The solutions begins with:

Viewing $\mathbb{R}P^2$ as $D^2/{\sim}$ where we identify $p \in S^1 =
 \partial D^2$ with $−p$, we set $U$ to be a small disk around $0 \in 
 D^2$ and $V = \mathbb{R}P^2 − \{0\}$. Thus $U$ is contractible and $V$
deformation retracts onto the subspace $S^1/{\sim}$, which is
homeomorphic to $S^1$ through a loop joining $p$ and $−p$.
We consider the Mayer-Vietoris sequence of the cover $\mathbb{R}P^2 \times U, \mathbb{R}P^2 \times V$. Note that the intersection is given by $\mathbb{R}P^2 \times (U \cap V )$. As $U$ is
contractible $\mathbb{R}P^2 \times U$ is homotopy equivalent to $\mathbb{R}P^2$.

That's all fine. But my issue begins now by plugging in the cohomologies into the Mayer-Vietoris sequence which is given by
$$\cdots \to H^1(X;\mathbb{Z}) \to H^1(A;\mathbb{Z})\oplus H^1(B;\mathbb{Z}) \to H^1(A\cap B;\mathbb{Z}) \to H^2(X;\mathbb{Z})\to \cdots$$
Let's ignore the degree $0$ stuff for now.
I know that $$H^*(\mathbb{R}P^2,\mathbb{Z}) = \begin{cases} \mathbb{Z}/2\mathbb{Z} & * = 2 \\ 0 & \text{else} \end{cases}$$
and clearly $$H^*(S^1,\mathbb{Z}) = \begin{cases} \mathbb{Z} & * = 1 \\ 0 & \text{else} \end{cases}$$
Since $A:= \mathbb{R}P^2 \times U$ is homotopy equivalent to $\mathbb{R}P^2$, we have $$H^*(A;\mathbb{Z}) = H^*(\mathbb{R}P^2\times U;\mathbb{Z}) \cong H^*(\mathbb{R}P^2;\mathbb{Z})$$
$B:= \mathbb{R}P^2 \times V$ is homotopy equivalent to $\mathbb{R}P^2\times S^1$ and since $S^1$ is finitely generated free abelian, by the Künneth formula we get
$$ H^*(\mathbb{R}P^2\times V;\mathbb{Z}) \cong H^*(\mathbb{R}P^2\times S^1;\mathbb{Z}) = H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z})$$

Here is where i have issues:
Since $H^n(\mathbb{R}P^2;\mathbb{Z})$ vanishes in degree other than $n=2$ and $H^*(S^1;\mathbb{Z})$ vanishes in degree other than $n=1$, the tensor product $H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z})$ should give me always $0$, shouldn't it?
Somehow this seems to be wrong but i don't know why.
 A: 
$H^n(\mathbb{R}P^2;\mathbb{Z})$ vanishes in degree other than $n=2$ and $H^*(S^1;\mathbb{Z})$ vanishes in degree other than $n=1$

This is only true for reduced cohomology.  For unreduced cohomology, both of them have $\mathbb{Z}$ in $H^0$ as well.
In addition, note that $H^*(\mathbb{R}P^2;\mathbb{Z}) \otimes H^*(S^1;\mathbb{Z})$ does not mean you are taking the tensor product $H^n(\mathbb{R}P^2;\mathbb{Z}) \otimes H^n(S^1;\mathbb{Z})$ separately in each degree.  Rather, it means you are taking the tensor product of the entire cohomology rings (all degrees together).  This means that the degree $n$ part of the tensor product will be the direct sum of all ways to tensor parts whose degrees add up to $n$, i.e. $\bigoplus_{p+q=n}H^p(\mathbb{R}P^2;\mathbb{Z}) \otimes H^q(S^1;\mathbb{Z})$.  So for instance, the nontrivial $H^2(\mathbb{R}P^2;\mathbb{Z})$ and $H^1(S^1;\mathbb{Z})$ will get tensored together to give something nontrivial in $H^3$.  (Note that this additionally means that even if you don't care about $H^0$ in the end, you still have to take into account the unreduced $H^0$ of each factor, since terms like $H^0\otimes H^n$ and $H^n\otimes H^0$ will appear in the $n$th degree of the tensor product.)
