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.
Best Answer
There are a lot of different approaches depending upon what you know, what precise incarnation of cohomology you're using, and so forth.
Off the top of my head, the following seems easiest: for any $n \in \mathbb{Z}^+$, complex projective $n$-space $\mathbb{C} \mathbb{P}^n$ has the natural structure of a CW-complex with exactly one cell in each degree $2k$ for $0 \leq k \leq n$ and no cells in any other degrees. (You can see this by writing projective $n$-space as affine $n$-space union projective $n-1$ space.) This makes the co/homology work out about as simply as it gets: all the attaching maps are zero, so we have $H_i(\mathbb{C} \mathbb{P}^n, \mathbb{Z})$ is $\mathbb{Z}$ if $0 \leq i \leq 2n$ and $i$ is even and $0$ otherwise. The vanishing of all the odd homology groups also makes the Universal Coefficient Theorem easy to apply, and one gets exactly the same answer for the cohomology groups.
If you wanted to use some other method -- or, if you need to know the ring structure of the cohomology (short answer: it is a truncated polynomial ring) -- then please let me know.