Closed unit ball of an infinite-dimensional Banach space is not compact

I have a few questions about the following proof taken from https://math.berkeley.edu/~sarason/Class_Webpages/solutions_202B_assign11.pdf.

Prove that the closed unit ball of an infinite-dimensional Banach space is not compact.

Proof. Let $$B$$ be an infinite-dimensional Banach space. Choose any unit vector $$x_1\in B$$, and let $$A_1$$ be the subspace spanned by $$x_1$$. In the quotient space $$B/A_1$$, choose a coset of norm $$\frac{1}{2}$$ and then a representative $$x_2$$ of that coset of norm at most 1. Then $$\|x_2−x_1\| \geq \frac{1}{2}$$. Let $$A_2$$ be the subspace spanned by $$x_1$$ and $$x_2$$,and note that $$A_2$$ is closed. In the quotient space $$B/A_2$$, choose a coset of norm $$\frac{1}{2}$$ and then a representative $$x_3$$ of that coset of norm at most 1. Then $$\|x_3−x_2\| \geq \frac{1}{2}$$ and $$\|x_3−x_1 \| \geq \frac{1}{2}$$. Continuing in this way, we obtain a sequence $$\{x_n\}$$ of vectors in the closed unit ball of $$B$$ such that $$\|x_m−x_n\| \geq \frac{1}{2}$$ whenever $$m\neq n$$. The sequence $$\{x_n\}$$ then has no convergent subsequences, implying that the closed unit ball in $$B$$ is not compact.

My questions are:

1) How do we know that a coset with a norm $$\frac{1}{2}$$ exists in $$B/A_1$$?

2) Why does the closedness of $$A_2$$ matter?

3) Why does it matter if $$B$$ is a Banach space?

1. Because $$B / A_1$$ is a normed space, and it is not the zero space unless $$A_1 = B$$, which is not the case here ($$B$$ is infinite-dimensional and $$A_1$$ is one-dimensional). So it contains at least one nonzero element (remember elements of $$B / A_1$$ are cosets), and you can multiply this by an appropriate scalar to obtain something of norm $$1/2$$.
2. If you take a quotient of a normed space by a non-closed subspace, in general you only get a seminormed space (the resulting "norm" may fail to be positive definite). Since we want $$B / A_2$$ to actually be a normed space (see step 1), we need $$A_2$$ to be closed.