Let $A \subset X$ a set of all characteristic functions of finite sets. Show that no sequence in $A$ converges to the constant map $c(x)=1$.

Break it down, using the definitions:

$X$ is

just the collection of functions from $\mathbb R$ to $\mathbb R.$

A $sequence$ in $X$ is then

a collection $(f_n)_{n\in \mathbb N}:f_n\in X.$

If $f_n\in A\subset X$, then $f_n(x)=0$

for all but finitely many values of $x.$

Therefore $\{x: \exists n\in \mathbb N( f_n(x)=1)\}$

is countable.

If $f_n\to c$ then for each $x\in \mathbb R,$

$f_n(x)=1$ if $n$ is large enough,

which is impossible.

edit: it may be easier to see this using the ordered pair definition of the $f_n.$ Each $f_n$ is a set whose members are either $\langle x,0\rangle$ or $\langle x,1\rangle$ and the number of the latter is finite. Taking the union over $n$, it follows that the total number of elements of the form $\langle x,1\rangle$ in the collection $\{f_n\}_n$ is at most countable. If the $f_n$ were to converge to the constant function $c=1$ then, for each real number $x$ some member of $f_n$ would have to contain an element of the form $\langle x,1\rangle.$ This is a contradiction, since $\mathbb R$ is uncountable.