Let $B=\operatorname{End}_R\left(R^{(\mathbb N)}\right)$. Define $u,v \in B$ as $$u(e_{2_{i+1}})=0,\, u(e_{2_i})=e_i\\v(e_{2_{i+1}})=e_i,\,v(e_{2_i})=0$$
Prove that $\{u,v\}$ is a basis of $B$ as a $B$-module.
I've already checked that $B$ is indeed a ring with the operations:
$(f+g)(h)(n)=f(h)(n)+f(g)(n), \,(f\cdot g)(h)(n)=f(h)(n)\cdot g(h)(n)$.
I am having problems proving that any element in $\operatorname{End}_R\left(R^{(\mathbb N)}\right)$ can be written as a linear combination of $u$ and $v$, i.e. that if $f \in \operatorname{End}_R\left(R^{(\mathbb N)}\right)$, then there are $g_u,g_v \in \operatorname{End}_R\left(R^{(\mathbb N)}\right)$ such that $$f=g_u\cdot u+g_v\cdot v$$ Since $\big\{\{e_i\} : i \in \mathbb N\big\}$ is a basis of $R^{(\mathbb N)}$ as an $R$-module, then $u$ and $v$ are defined by the values they take at each $e_i$. I don't see how I can write each element of $B$ as a linear combination of $u$ and $v$.
To show it is basis, I would also have to prove that the set $\{u,v\}$ is linearly independent. I've tried to write $0_B$, the neutral element of $B$ as a combination of $u,v$: $$0_B=\lambda\cdot u+\beta\cdot v.$$ I would like to conclude $\lambda=\beta=0_B$. I evaluate $0_B$ in $e_{2i+1}$, so $$0_{A^{(\mathbb N)}}=\lambda(e_{2i+1})\cdot u(e_{2i+1})+\beta(e_{2i+1})\cdot v(e_{2i+1})\\=\beta(e_{2i+1})\cdot e_i$$
Analogously one concludes $$\lambda(e_{2i})\cdot e_i=0_{A^{(\mathbb N)}}.$$
I couldn't deduce from here that $\lambda$ and $\beta$ are the zero element in $B$.
Any suggestions would be appreciated.
It seems you missed the multiplicative operation on $\operatorname{End}_R\left(R^{(\mathbb N)}\right)$, denoted by "$\cdot$". Actually this is always the composition of functions. Then $$0_{R^{(\mathbb N)}}=\lambda(u(e_{2i+1}))+\beta(v(e_{2i+1}))=\beta(e_i),$$ so $\beta=0$.