Neighbourhoods in topology 
Let $X=C[0,1]$ and consider the topology $\tau=\tau(S)$ generated by $$S=\{V_{x,U}\}_{x\in[0,1],~U=(a,b)\subset\Bbb R},$$ where $$V_{x,U}=\{f\in C[0,1]:f(x)\in U\}$$
$1)$ Let $V\in\tau$ be a neighborhood of the (constant) zero function $$f(x)=0,\forall x\in\Bbb R$$ Show that there is a finite set of points $x_1,\dots,x_n\in[0,1]$ such that $V$ contains all functions $g$ which satfisfy $$g(x_1)=g(x_2)=\dots=g(x_n)=0.$$
$2)$ Continuing the previous part, and with the same notation, suppose that $$x\in[0,1]\setminus\{x_1,\dots,x_n\}.$$
  Show that there is a function $h\in C[0,1]$ such that $h(x)=i$ and $h\in V$.


I think I'm having trouble understanding what it actually means for something to be a neighbourhood. In terms of metric spaces, it's just inside an open ball which makes sense, in a topology a neighbourhood is just any open set containing the point ( in this case a function ) right? 
i) So If $V$ is a neighbourhood of the constant zero function, then it has an open set containing f as well as other functions, is there anything special about all the other functions inside that open set? or is just any function as long as its continuous on $C[0,1]$ and restricted to the open interval U?
I guess if V is a neighborhood then any open set inside V would contain functions that are restricted to the interval around 0, but since f is the constant zero function that satisfies $f(x) = 0$ $ \forall x \in [0,1] $ (infinitely many points) then any other function can only have finitely many points $x_{1} , ...x_{n}$ that satisfy $g(x_{1}) , ... g(x_{n}) = 0$. Is this correct?
ii) From i) since any neighborhood contains functions with finitely many points that go to zero. Then we choose an $x$ that isnt the same as any of the finitely many points (which is possible since $[0,1]$ is uncountably infinite ) and find a continuous function $ h(x) = 1$ since there are no restrictions on the functions as long as $ x \in [0,1] \setminus \{x_{1} , .. x_{n} \} $
Is my understanding here correct? Not sure if its entirely correct.
Any clarification would be helpful.
 A: The only way you can test that $O$ is an open neighbourhood of the $0$-function, is that there exist finitely many $x_1, \ldots, x_n\in [0,1]$ and corresponding open intervals $U_1,\ldots, U_n$ such that $$0 \in V_{x_1, U_1} \cap \ldots \cap V_{x_n, U_n} \subseteq O$$
This is what it means that the topology on $C([0,1])$ is generated by the collection $S$ from your problem statement. (The finite intersections of subbasic elements form a base....)
Note that the fact that $0$ is in this finite intersection (which has value $0$ on any $x_i$ of course) is that $0 \in U= U_1 \cap U_2 \cap \ldots U_n$ and as $V_{x,U} \subseteq V_{x,U_i}$ for all $i$, we see that in fact (for $0$, which is an easier case) $O$ is an open neighbourhood of $0$ iff there are finitely many $x_1, x_2, \ldots, x_n$ and an open interval $U$ containing $0$, such that 
$$\bigcap_{i=1}^n V_{x_i,U} \subseteq O$$
So the only concrete information on $O$ we have from the definitions then is that it contains any $f \in C([0,1])$ that sends all $f(x_i)$ inside $U$. In particular, it contains all functions $g$ that obey $g(x_1) = g(x_2) = \ldots g(x_n)=0$, as $0 \in U$.
This finite set of points will differ for each open neighbourhood of $0$. But given those finitely many points belonging to some open neighbourhood $V$ (as in problem 2) if we have any other point $x \in [0,1]$, we can easily find a continuous piecewise linear function $h$ that has a peak at $x$ at height $1$ (rising from constantly $0$ and going back down linearly in a small interval around $x$ that misses all $x_1, \ldots, x_n$. This $h$ then is $0$ on all $x_i$ so $h \in V$ is garantueed (by the above) but also $h(x)=1$. 
A: For question 1. We may assume that $V$ is a basic neighborhood (if not, then replace it with a basic neighborhood). Since $S$ is a subbase of the topology (which is implied by the word "generates") then $V$ is the intersection of finitely many sets, each in $S$. Say  
$V=V(x_1,U_1)\cap V(x_2,U_2)\cap\dots\cap V(x_n,U_n)$. 
Since $f\in V(x_k,U_k)$ for each $k=1,...,n$ we must have that $f(x_k)\in U_k$. That is, $0\in U_k$ each $k=1,...,n$. So, if $g(x_1)=...=g(x_n)=0$ then $g(x_k)\in U_k$ for each $k$, hence $g\in V(x_k,U_k)$ and $g\in V$. 
