Showing theres no metric wrt a topology 
I have a homework problem that I have no clue how to approach and answer. My topology is kind of weak not entirely sure that I understand the question. 
My attempt:
i) If  $V_{1}$ is a sequence of neighborhoods of the zero function then there is a function say $h_{1} \in C[0,1] $ such that $h_{1} (x) = 1 $ where $h \in V$ [ not entirely sure why this is true but a friend told me this] . Now if we let sequence of neighbourhoods $ V_{1} , V_{2} , .. V_{n} $ to contain functions $h_{1} , h_{2} ... h_{n} $ respectively such that all $h_n(x) = 1$. Is this correct? 
ii) How do I go about showing that no metric exist? 
Any help is appreciated. 
 A: The answer given doesn't work, as it's not clear what the point $x$ or the functions $h_1, \ldots, h_n$ are. You don't seem to have constructed anything here.
Note that each neighbourhood only controls the functions within it at a single point. If we consider, for example, $V_{1, (3,4)}$, the functions in this neighbourhood are precisely the continuous functions $f : [0, 1] \to \Bbb{R}$ such that $3 < f(1) < 4$. This poses no restriction on $f(x)$ for any $x \neq 1$; basically I can pick any $x \neq 1$ and any $y \in \Bbb{R}$, and find a function in $V_{1, (3,4)}$ such that $f(x) = y$.
We have a sequence of neighbourhoods $V_n$. Each of these neighbourhoods must contain a sub-basic neighbourhood $V_{x_n, U_n}$, by definition of the topology. Now, there are only countably many points in the sequence $(x_n)$, and $[0, 1]$ is uncountable, so there must be a point $x \in [0, 1]$ such that $x \neq x_n$ for all $n$. By the above argument, we can find a continuous function $h_n$ in each $V_n$ so that $h_n(x) = 1$, simply because $x \neq x_n$ for all $n$ (for example, take the equation for the line between points $(x, 1)$ and $(x_n, u_n)$, where $u_n$ is some point - your choice - in $U_n$).
For the second part, we need to observe a property of metric spaces that this topological space does not possess, probably to do with the previous part. In this case, it looks like the property of first countability might be an issue (why else would we consider a sequence of neighbourhoods of a single point?). Metric spaces have first countability because of the local basis $B_{1/n}(x)$ of open balls around each point $p$, so let's start by assuming a metric does exist, and applying the previous part to the sequence of open sets $B_{1/n}(0)$ for $n \in \Bbb{N}$.
By the previous part, there exists some $x \in [0, 1]$ and functions $h_n$ such that $h_n \in B_{1/n}(0)$ and $h_n(x) = 1$. In particular, note that $d(h_n, 0) < 1/n$ for all $n$, hence $h_n \to 0$.
Now, look at a neighbourhood $V = V_{x, (-1, 1)}$ of the $0$ function, and note that $h_n \notin V$ for any $n$. This contradicts $h_n \to 0$. Thus, by contradiction, no metric can exist for this topology.
A: Consider the set $Y=\Bbb R^{[0,1]}$ of all functions from $[0,1]$ to $\Bbb R,$ which can be considered to be the product of $2^{\aleph_0}$ copies of $\Bbb R,$ although we use $[0,1]$ for the index of the product.  A base $B$ for the (Tychonoff ) product topology on $Y$ is the set of all  $\prod_{x\in [0,1]}A_x$ such that
$(i). \{x:A_x\ne \Bbb R\}$ is finite, and
$(ii).$ If $A_x\ne \Bbb R$ then $A_x$ is a bounded open real interval.
Now $S$ is a sub-base for $\tau.$ The set $T$ of all $\cap W,$ over all $finite$ $W\subset S,$ is a base for $\tau.$ And $$T=\{b\cap C[0,1]: b\in B\}.$$ So $X$ with the topology $\tau$ is a sub-space of $Y$ with the Tychonoff product topology on $Y.$
Q i). Let $\{V_n\}_{n\in \Bbb N}$ be a sequence of nbhds of $any$ $f\in X.$ For each $n\in \Bbb N$ there exists $V'_n\in B$ such that $f\in V'_n\cap X\subset V_n.$
Let $V'_n=X\cap \prod_{x\in [0,1]}A_{x,n}$ where $\prod_{x\in [0,1]}A_{n,x}\in B.$ 
Let $C_n=\{x\in [0,1]: A_{x,n}\ne \Bbb R\}.$
Each $C_n$ is finite so $[0,1]\ne \cup_{n\in \Bbb N}C_n.$ So let $x_0 \in [0,1]\setminus \cup_{n\in \Bbb N}C_n.$
Now for each $n$, since $C_n$ is finite, and $x_0\not \in C_n ,$  and $A_{x_0,n}=\Bbb R,$ there exists $h_n\in X $ such that $h_n(x)=f(x)$ for all $x\in C_n$ and $h_n(x_0)=1+f(x_0).$
So for all $n\in \Bbb N$ we have $h_n\in V_n$ and $h_n(x_0)=1+f(x_0).$
Q ii). Suppose the $\tau$ topology is metrizable. Then it is first-countable: For every $f\in X$ there exists a countable set $\{V_n\}_{n\in \Bbb N}$ of nbhds of $f$ such that if $V$ is any nbhd of $f$ then there exists $n \in \Bbb N$ such that $V_n\subset V.$
But from Q i), consider the nbhd $V=C[0,1]\cap \prod_{x\in [0,1]}A'_x$ where $A'_x=\Bbb R$ if $x\ne x_0$ and $A'_{x_0}= (f(x_0)-1/2,\, f(x_0)+1/2)\,).$  For every $n\in \Bbb N$ we have $h_n\in V_n\setminus V$ so $V_n\not \subset V.$
Remark. In Q ii), if $d$ is a metric for a space $X$ and $f\in X$ we can let $\{V_n\}_{n\in \Bbb N}$ be the set $\{B_d(f,1/n):n\in \Bbb N\}$ of open balls. 
A: In this question you ask for the fact that if $V$ is a neighbourhood of $0$ (the function in $C([0,1])$) there is a finite set $F(V)$ such that if $f$ is $0$ on all points of $F(V)$ then $f \in V$. This is shown there in my answer (part 1).
So find such a finite set $F(V_n)$ for each $V_n$ when we are given the sequence $(V_n)$ of open neighbourhoods of $0$.
Then we can find $x \in [0,1]$ and such that $x \notin F(V_n)$ for every $n$. The union of the $F(V_n)$ is countable and $[0,1]$ is uncountable.
Now apply fact 2 from the above question for each $n$ and $V_n$: We know $x \notin F(V_n)$, so there is a function $h_n \in V$ with $h_n(x)=1$ (we can make a piecewise linear function that is $0$ on the finite set $F(V_n)$, ensuring $h_n \in V_n$, and $1$ on $x$ very easily). 
So i) is a straightforward application of the linked question using only that $[0,1]$ is uncountable as an extra step. 
Note that $h_n \not\to 0$ because it does not converge pointwise at $x$, or note that $V_{1,(-\frac12,\frac12)}$ is an open neighbourhood of $0$ that misses all $h_n$.
Now (ii) follows from the following observation which holds in any metric space $(X,d)$:

For every $x \in X$ there is a sequence of neighbourhoods $(V_n(x))$ of $x$, such that if we pick any $x_n \in V_n(x)$, then $x_n \to x$.

Proof: take $V_n = B(x, \frac1n)$ and check the easy details.
Now we have just shown under (i) that 

For $X = C([0,1])$ and $x=0$, whatever sequence of neighbourhoods $(V_n)$ we choose of $0$, there always is a choice $h_n \in V_n$ such that $h_n \not \to 0$.

This shows that $C([0,1])$ is topologically different in an essential way from any metric space. So it's not metrisable. The point $0$ is coincidental BTW, this could be done for any function $f \in C[0,1])$, as the space is homogeneous. 
