The set of functions from $[0,1)$ to $\{0,1\}$ is not metrizable. 
Set $F = \{ f : f: [0,1) \to \{0,1\}\}$ and equip F with the topology of pointwise convergence. Where $\{0,1\}$ is equipped with the discrete topology.  Show this is not metrizable. 

I tried to use this question : C(X) with pointwise is not metrizable
But I am slightly confused. I understand that if we can show that it is not first countable then it is not metrizable. 
In the answer to the above question it states: 
" By definition of the topology of pointwise convergence, for each $i$ there exist some finite set $J_i$ and some $\epsilon_i > 0$ such that $f \in U_i$ if $|f(x)| < \epsilon_i$ for all $x \in J_i$.  Now define another neighbourhood $U$ of $0$, and use the fact that $X$ is completely regular to show that for any $i$ there is some $f \in C(X)$ such that $f \in U_i$ but $f \notin U$. "
I am not clear on why this finite set $J_i$ exists. 
Maybe an explanation of that would help me understand what to do.
 A: In the general case, looking at some family $F$ of functions $X\rightarrow Y$, the topology in question is the subspace topology of $F\subset Y^X$ inherited from the product topology, as discussed in this question. Now the product topology on $Y^X$ is generated by products of open sets, $\prod_{x\in X}U_x$, where $U_x=Y$ for $\textit{all but finitely many }x$. 
Thus, if we pick $\ f\in F$ and any open set $\mathcal{U}\ni f$, we have that $\mathcal{U} = F\cap \widetilde{\mathcal{U}}$ for some open set $\widetilde{\mathcal{U}}\subset Y^X$, where $\widetilde{\mathcal{U}}$ must contain some product set $\widetilde{\Pi}$ like in the above paragraph. Thus we also have $\Pi= F\cap \widetilde{\Pi}\subset \mathcal{U}$. But since all but finitely many factors in the product set are $Y$, the condition for $g\in \Pi$ is simply that $g(x_i)\in U_{x_i}$ for some finite set of points $x_1,\ldots, x_n$.
Now specifying to you case, we would have $X=[0,1)$ and $Y=\{0,1\}$, and $U_i$ is some open set in $F$ containing the zero function. By our above reasoning, $f\in U_i$ is implied by $f(x_j)\in U_{x_j}$ for each point in a finite set $x_1,\ldots,x_n$, each $\ U_{x_j}$ containing zero. Furthermore, since $Y$ is a metric space, $f(x_j)\in U_{x_j}$ is implied by $|f(x_j)|<\varepsilon_{x_j}$ for some $\varepsilon_{x_j}>0$. Now take $\varepsilon_i = \text{min}\{\varepsilon_{x_1},\ldots,\varepsilon_{x_n}\}$ and $J_i = \{x_1,\ldots,x_n\}$ in what you quoted in your question.
A: You only need to understand what is exactly the topology of pointwise convergence. As you may know, we can distinguish a topology from the others once we have an explicit description of: open sets, or closed sets, or a basis of neighborhoods for each point of the space.
For the topology of pointwise convergence, in particular, my favorite way to define it is using an explicit description of a basis of neighborhoods for each point of the space. This is done as follows:
For $f\in F$, the collection of all the sets of the form $B_{J}(f,\varepsilon)$ constitute a basis of neighborhoods for $f$, where $\varepsilon\in\mathbb{R}^+$, $J$ is a finite subset of $[0,1)$ and 
$$B_{J}(f,\varepsilon):=\{g\in F:|g(x)-f(x)|<\varepsilon,\mbox{ for all }x\in J\}.$$
Now, why $J$ has to be finite? Well, the following alternative (but equivalent) definition of the topology of pointwise convergence is going to give you the answer.
For each $x\in [0,1)$, define the seminorm $p_x:F\to\mathbb{R}$ as $p_x(f)=|f(x)|$. The topology of pointwise convergence is defined as the coarsest locally convex topology on $F$ such that all the seminorms $p_x$ are continuous.
You can prove that all the sets of the form $B_{\{x\}}(f,\varepsilon)$ constitute a subbase for this topology. If we consider all the possible finite intersection of members of a subbase we obtain a base for the topology.
Hence sets of the form 
$$B_{\{x_1\}}(f,\varepsilon_1)\cap B_{\{x_2\}}(f,\varepsilon_2)\cap\dots\cap B_{\{x_n\}}(f,\varepsilon_n)=B_{J}(f,\varepsilon)$$
where $J=\{x_1,\dots,x_n\}$ and $\varepsilon=\min\{\varepsilon_1,\dots,\varepsilon_n\}$, constitute a base for the topology of pointwise convergence.
