If you remove a point from $P^2$ you are left with something which looks like a Moebius band. You can use this to compute $H^\bullet(P^2)$.
Let $p\in P^2$, let $U$ be a small open neighborhood of $p$ in $P^2$ diffeomorphic to an open disc centered at $p$, and let $V=P^2\setminus\{p\}$. Now use Mayer-Vietoris.
The cohomology of $U$ you know. The open set $V$ is diffeomorphic to an open moebious band, so that tells you the cohomology; alternatively, you can check that it deformation-retracts to the $P^1\subseteq P^2$ consiting of all lines orthogonal to the line corresponding to $p$ (with respect to any inner product in the vector space $\mathbb R^3$ you used to construct $P^2$), and the intersection $U\cap V$ has also the homotopy type of a circle. The maps in the M-V long exact sequence are not hard to make explicit; it does help to keep in mind the geometric interpretation of $U$ and $V$.
Later: alternatively, one can do a bit of magic. Since there is a covering $S^2\to P^2$ with $2$ sheets, we know that the Euler characteristics of $S^2$ and $P^2$ are related by $\chi(S^2)=2\chi(P^2)$. Since $\chi(S^2)=2$, we conclude that $\chi(P^2)=1$. Since $P^2$ is of dimension $2$, we have $\dim H^p(P^2)=0$ if $p>2$; since $P^2$ is non-orientable, $H^2(P^2)=0$; finally, since $P^2$ is connected, $H^0(P^2)\cong\mathbb R$. It follows that $1=\chi(P^2)=\dim H^0(P^2)-\dim H^1(P^2)=1-\dim H^1(P^2)$, so that $H^1(P^2)=0$.
Even later: if one is willing to use magic, there is lot of fun one can have. For example: if a finite group $G$ acts properly discontinuously on a manifold $M$, then the cohomology of the quotient $M/G$ is the subset $H^\bullet(M)^G$ of the cohomology $H^\bullet(M)$ fixed by the natural action of $G$. In this case, if we set $M=S^2$, $G=\mathbb Z_2$ acting on $M$ so that the non-identity element is the antipodal map, so that $M/G=P^2$: we get that $H^\bullet(P^2)=H^\bullet(S^2)^G$.
We have to compute the fixed spaces:
$H^0(S^2)$ is one dimensional, spanned by the constant function $1$, which is obviously fixed by $G$, so $H^0(P^2)\cong H^0(S^2)^G=H^0(S^2)=\mathbb R$.
On the other hand, $H^2(S^2)\cong\mathbb R$, spanned by any volume form on the sphere; since the action of the non-trivial element of $G$ reverses the orientation, we see that it acts as multiplication by $-1$ on $H^2(S^2)$ and therefore $H^2(P^2)\cong H^2(S^2)^G=0$.
Finally, if $p\not\in\{0,2\}$, then $H^p(S^2)=0$, so that obviously $H^p(P^2)\cong H^p(S^2)^G=0$.
Luckily, this agrees with the previous two computations.
Your result isn't correct.
I won't tell you the result so that you can compute it yourself, though :) There are very few things more rewarding than getting these things right oneself!
(By the way: do it for $k=1$, $2$, and $3$... you'll catch the pattern soon)
Best Answer
You can take these two open sets (in blue and red) :
The long exact Mayer-Vietoris sequence is $$0\rightarrow H^0(M)\rightarrow H^0(U)\oplus H^0(V)\rightarrow H^0(U\cap V)\rightarrow H^1(M)\rightarrow H^1(U)\oplus H^1(V)\rightarrow$$
But the Möbius strip is connected, so $H^0(M)=\mathbb R$.
The open sets are connected, so $H^0(U)=H^0(V)=\mathbb R$.
The intersection has two connected components so $H^0(U\cap V)=\mathbb R^2$.
The opens sets are contractible so $H^1(U)=H^1(V)=0$.
So you have the exact sequence $$0\rightarrow\mathbb R\rightarrow \mathbb R^2\rightarrow\mathbb R^2\rightarrow H^1(M)\rightarrow0$$
So you have $1-2+2-\dim H^1(M)=0\Rightarrow \dim H^1(M)=1$.
Next, $$H^1(U)\oplus H^1(V)=0\rightarrow H^1(U\cap V)=0\rightarrow H^2(M)\rightarrow H^2(U)\oplus H^2(V)=0$$ gives you $H^2(M)=0$.
Conclusion: $H^k(M)=\left\{\begin{array}{ll}\mathbb R & \text{ if }k=0,1 \\ 0 & \text{ else}\end{array}\right.$
Remark: without the MV sequence, you could notice that the median circle $\mathbb S^1$ is a deformation retract of the Möbius strip and use https://math.stackexchange.com/a/162378/33615.