Is construction of countable sequence is sufficient to prove following theorem? 
$R^\omega$ with box topology is not metrisable.

I understand almost whole proof. given as below

I do not understand the following 
The author created all possible sequence in $R^\omega$ and then use diagonalization argument to arrive at the contradiction
But Problem is that with this construction only countable number of sequences possible .
But which is not case as $R^\omega$ is uncountable.W e can find some sequence which is not in construction.
Maybe my doubt is silly . But I couldn't understand.
Any Help will be appreciated
 A: The $(a_n)$ sequence is taken arbitrarly from $A$, there is no construction here. And yes, he goes over all sequences, but he never states (or indirectly uses a fact) that there are countably many of them. The diagonalization argument is there to show that $(a_n)$ cannot converge to $0$. And since it was chosen arbitrarly from $A$ then $\mathbb{R}^\omega$ is not metrizable because $0\in\overline{A}$. 
Perhaps your confusion comes from the fact that $(a_n)$ is actually a sequence of sequences? He uses this weird double index notation, we would write it like this:
$$\begin{matrix}
a_1 & = & (a_{11}, a_{12}, a_{13}, a_{14}, \ldots) \\
a_2 & = & (a_{21}, a_{22}, a_{23}, a_{24}, \ldots) \\
a_3 & = & (a_{31}, a_{32}, a_{33}, a_{34}, \ldots) \\
a_4 & = & (a_{41}, a_{42}, a_{43}, a_{44}, \ldots) \\
\vdots & & \vdots \\
a_n & = & (a_{n1}, a_{n2}, a_{n3}, a_{n4}, \ldots) \\
\vdots & & \vdots
\end{matrix}$$
and when he says that $(a_n)$ converges to $0$ he actually means $0=(0,0,0,\ldots)$ as an element of $\mathbb{R}^\omega$.
