I have obtained that the cohomology rings
$$
H^\*(G\_k(\mathbb{R}^\infty);\mathbb{Z}\_2)=\mathbb{Z}\_2[w\_1,\cdots,w\_k].
$$
Also 
$$
H^\*(G\_k(\mathbb{R}^m);\mathbb{Z}\_2)=\mathbb{Z}\_2[w\_1,\cdots,w\_k]/(\bar w\_{m-k+1},\cdots,\bar w\_{m}).
$$
The inclusion $i: G\_k(\mathbb{R}^m)\to G\_k(\mathbb{R}^\infty)$ induces a natural quotient map in the above cohomology algebras. 


What are the cohomology rings with integral or rational coefficients


$$
H^\*(G\_k(\mathbb{R}^\infty);\mathbb{Q})?
$$
$$
H^\*(G\_k(\mathbb{R}^m);\mathbb{Q})?
$$
$$
H^\*(G\_k(\mathbb{R}^\infty);\mathbb{Z})?
$$
$$
H^\*(G\_k(\mathbb{R}^m);\mathbb{Z})?
$$



integral or rational cohomology of real grassmannians
This is just answer in the very special case of integer cohomology of $\mathbb{R}\mathbb{P}^\infty$ and $\mathbb{R}\mathbb{P}^n$ as asked in one of the comments.


$\mathbb{R}\mathbb{P}^\infty$ can be divided into cells $\mathbb{R}\mathbb{P}^k\backslash
\mathbb{R}\mathbb{P}^{k-1}=\mathbb{R}^k$, namely one cell in each dimension. The differentials are either 0 or multiplication by 2, depending on parity of the dimension. The cell complex computing integer cohomology is
$$\mathbb{Z}\overset{0}{\to}\mathbb{Z}\overset{2}{\to}\mathbb{Z}\overset{0}{\to}\mathbb{Z}\overset{2}{\to}\dots$$
Hence the integer cohomology groups are
$$H^0=\mathbb{Z}, H^1=0,H^2=\mathbb{Z}/2\mathbb{Z}, H^3=0,H^4=\mathbb{Z}/2\mathbb{Z},\dots.$$


For the space $\mathbb{R}\mathbb{P}^n$ the above complex should be truncated at degree $n$. Then it is equally easy to compute its cohomology.


In general, real Grassmannians can be decomposed into real Schubert cells. But I do not know whether the corresponding cell complex can be used to compute the cohomology.


