I want to know whether my proof is correct:
I claim that the sequence of indicator functions $x_n = 1_{A_n}$ (where $A_n = \{n\}$) does not converge in $\{0,1\}^{\mathbb{R}}$ and has no convergent subsequence. Indeed, if $A \subset \mathbb{R}$ is non-empty then there exists $x \in A$, so the basic open set defined by $B = \displaystyle{\prod_{\alpha \in \mathbb{R}} U_{\alpha}}$ where $U_\alpha = \{1\}$ if $\alpha = x$ or $\alpha = c(x+1)\doteq d$ (where $c(x)$ stands for the ceiling function) and $U_\alpha = \{0,1\}$ everyhere else obviously contains no tail of the sequence $(x_n)$ (since $1_{A_n}(d) = 0 $ for all $n > d$ and so $1_{A_n} \notin B$ for all $n > d$ -it doesn't contain one tail, therefore it doesn't contain any other-). An almost identical argument applies to show that it no subsequence converges.
Is everything okay? Could I improve anything here?
EDIT: it's all wrong. I'll try to fix it.