Limit of an indicator function?

Answering your first question: For a given $x$, there are at most two values of $n$ for which $1_{[n,n+1]}$ is nonzero. As $n\to\infty$, the interval which supports $1_{[n,n+1]}$ will "fly off" to the right, passing any given $x$. So in the tail of the sequence, $1_{[n,n+1]}(x)$ is eventually zero for any give $x$.