I have several doubts on attacking the following topology questions I am doing some exercise related to topology on the real line. The excercise (translated from French) goes like:
Given $E = \mathcal{F}([0,1], \mathbb{R})$ the set of real functions on $[0,1]$

*

*Show that for all $x \in \mathcal{F}([0,1], \mathbb{R}) $ the familly $\mathcal{B}(x)$ composed of  subsets $B_x$ of $\mathcal{F}$:
$$ \mathcal{B}_x = \{B_x = B(x; t_1, t_2, ..., t_n; \epsilon)\}= \{y= y(t) \in E : |y(t_i) - x(t_i)| < \epsilon, n \in  \mathbb{N}\} \\ y \in B_x \in \mathcal{B}_x \iff \exists  \mathbb{N}, \exists \epsilon > 0 : y\in B(x; t_1, t_2, ..., t_n; \epsilon) \}$$
is a base of neighborhoods of $x$ in $E $


*Let $A$ be the set of functions of $\mathcal{F}$  everywhere equal to one except in a finite number where they are zero:
$$ A = \{ x \in E \mid \exists n \in \mathbb{N}, \exists t_1, ... t_n \in [0,1] : x(t_i)= 0 ,  x(t) = 1\ \forall  t \neq t_i, I = 1,2,...,n  \} $$
i. Show that in $\mathcal{F}$ with the topology defined in 1), the function $x_0 \equiv 0$ on $[0,1]$ is an accumulation point of $A$. 
ii. Show that for any sequence $(x_n(t))_n$ in $A$, the set of all points where at least one of the functions of this self-canceling sequence is countable. Deduce that there exists $t_0 \in [0,1]$ such that $x_n(t_0) \neq 0 , \forall n \geq 1$ . 
iii. Show that the neighborhood $B(0; t_0; \frac{1}{2})$ does not contain any term of this sequence. 
iv. Deduce that the adherence $\overline{A}$ cannot be obtained by adding to $A$ the limits points of $A$ (hint. the preceding sequence $(x_n)_n$ does not converge to $0$).
What I tried:
The first question is relatively easy I proved it my using properties of a basis of a topology.
From question 2 is where I get troubles, for part i) I think that every neighborhood of the function $x_0 \equiv 0$ containing that function necessarily contains the function $x(t) = 1$ as defined in set $A$. (have to put it in symbolic form) 
For part ii) I think I need to construct some bijective function of some sort, whose image is zero for all $\forall  t = t_i$.
I am a beginner in topology and I would appreciate hints on attacking the questions
 A: (i) Consider any basic neighborhood $B_0(t_1, \ldots, t_n; \epsilon)$ of $x_0 \equiv 0$ for some $t_1, \ldots, t_n \in [0, 1]$ and some $\epsilon > 0$. You just have to come up with a function in $A$ that lies in $B_0(t_1, \ldots, t_n; \epsilon)$. But this is easy. Just define
$$
x_{t_1, \ldots, t_n}(t) := \begin{cases} 1 &\text{if } t \in [0, 1] \setminus \{t_1, \ldots, t_n\} \\ 0 &\text{if } t \in \{t_1, \ldots, t_n\} \end{cases}
$$ Obviously $x_{t_1, \ldots, t_n} \in B_0(t_1, \ldots, t_n; \epsilon)$ since $|x_0(t_i) - x_{t_1, \ldots, t_n}(t_i)| = |0 - 0| = 0 < \epsilon$ for $i \in \{1, \ldots, n\}$. And $x_{t_1, \ldots, t_n} \in A$ because by definition it is equal to $1$ except at a finite number of points $t_1, \ldots, t_n$.
In other words, we have just shown that any basic neighborhood of $x_0 \equiv 0$ no matter how small contains an element of $A$ and therefore $x_0$ must be an accumulation point of $A$.
(ii) Let $(x_n)_{n \in \Bbb N}$ be a sequence in $A$. By definition of $A$, this means that for each $n \in \Bbb N$, $x_n$ is equal to $1$ except at a finite number of points $t_{1}^n, \ldots, t_{m(n)}^n \in [0, 1]$ (for some $m(n) \in \Bbb N$) where $x_n$ is $0$.
Thus, if we let $X_0$ be the collection of all points where where at least one of the functions in the sequence is $0$, then clearly we must have
$$
X_0 := \big\{t \in [0, 1]\ \big|\ \exists n \in \Bbb N : x_n(t) = 0\big\} = \bigcup_{n \in \Bbb N} \big\{t_{1}^n, \ldots, t_{m(n)}^n\big\}
$$ This is a countable set by a standard result from set theory: a countable union of finite sets is at most countable.
