Show that the set $A = \{n \alpha \pmod 1\mid n ∈ \mathbb Z\}$ is countably infinite, where $\alpha$ is an irrational number. I'm trying to understand a solution to a counting problem. I would like to receive detailed explanations to the different drawn conclusions in the solution, thank you.
Q: Given $x ∈ R $, let $y = x \pmod 1$, where $y ∈ [0,1)$, such that $x = n + y$ for some $n ∈ Z$. Now let $\alpha$ be an irrational number, which finally gives us the set $A = [n \alpha (mod 1)| n ∈ Z]$. Show that the set A is countably infinite.
S: Note that $A < B = [n \alpha - k| n,k ∈ Z]$. Now in the solution it is apparently obvious that the set $B$ is countable since it is a countable union of countable set, however I do not see that. If $B$ is a union what would each countable set $B_i$ look like?
Continuing the solution says: if we can show that $A$ contains infinitely many elements then we have proved that $A$ is countably infinite. Note that  if $y_i ∈ A$ and $m_i ∈ Z$ for $i = 1, 2$ and $m_1 y_1 + m_2 y_2 ∈ [0,1)$, then $m_1 y_1 + m_2 y_2 ∈ A$. Reason being that if $y_i ∈ A$ then we have $y = n_i \alpha - k_i$ for some $k_i$ and $n_i$. Hence we have $m_1 y_1 + m_2 y_2 = (m_1 n_1 + m_2 n_2) \alpha - m_1 k_1 - m_2 k_2$ and that $m_1 y_1 + m_2 y_2 ∈ [0,1)$, where $m_1 y_1 + m_2 y_2 = (m_1 n_1 + m_2 n_2) \alpha (mod 1)$, which gives the wanted conclusion.
Now, why is it necessary that $m_1 y_1 + m_2 y_2 ∈ A$ if $y_i ∈ A$ and $m_i ∈ Z$ for $i = 1, 2$ and $m_1 y_1 + m_2 y_2 ∈ [0,1)$? And further, how do I understand that $A$ is infinite from realizing $m_1 y_1 + m_2 y_2 = (m_1 n_1 + m_2 n_2) \alpha - m_1 k_1 - m_2 k_2$ and that $m_1 y_1 + m_2 y_2 ∈ [0,1)$, where $m_1 y_1 + m_2 y_2 = (m_1 n_1 + m_2 n_2) \alpha (mod 1)$?
I realize that the solution uses the theorem: "every infinite subset of a countable set $C$ is countable".
Thank you
 A: An explanation of what's going on

Why is $B = \{n \alpha - k : n, k \in \mathbb Z\}$ countable?

It is the union of the countable sets $\mathbb Z + n \alpha$ over all $n \in \mathbb Z$ (where by $\mathbb Z + n \alpha$ I mean adding $n\alpha$ to each element of $\mathbb Z$).

If $y_1, y_2 \in A$ and $m_1 y_1 + m_2 y_2 \in [0,1)$, why is $m_1 y_1 + m_2 y_2 \in A$?

We can write $y_1 = n_1 \alpha \bmod 1 = n_1 \alpha - k_1$ for some integers $n_1, k_1$, and similarly $y_2 = n_2 \alpha - k_2$ for some integers $n_2, k_2$. Therefore $$m_1 y_1 + m_2 y_2 = (m_1 n_1 + m_2 n_2) \alpha - (m_1 k_1 + m_2 k_2).$$ Since $m_1 y_1 + m_2 y_2$ has the form "$(m_1 n_1 + m_2 n_2) \alpha$ minus an integer", and it is in the range $[0,1)$, it must be $(m_1 n_1 + m_2 n_2) \alpha \bmod 1$. That's an element of $A$ by definition.

Why does this tell us that $A$ is infinite?

By itself, it doesn't. The property above would be true even if $\alpha$ is rational; but if $\alpha$ were rational, then $A$ would be finite.
We have to additionally use the irrationality of $\alpha$ at some point, and the proof you're quoting never does. Maybe there's another part you haven't quoted? If not, then you're definitely not missing something obvious; there's just a gap in the proof.
A better proof
Rather than try to guess at what the intended use of that claim about $m_1 y_1 + m_2 y_2$ is, here is a shorter argument that avoids it entirely.
Define a function $f : \mathbb Z \to A$ by $f(n) = n\alpha \bmod 1$. It is surjective, because $A$ is defined as the image of $f$.
Lemma. $f$ is also injective: if $n_1, n_2 \in \mathbb Z$ with $n_1 \ne n_2$, then $(n_1 \alpha \bmod 1) \ne (n_2 \alpha \bmod 1)$.
Proof. We get $n_1 \alpha \bmod 1$ from $n_1 \alpha$ by adding some integer $m_1$ to get a value in the range $[0,1)$. This means that $(n_1 \alpha \bmod 1) = n_1 \alpha + m_1$. Similarly, there is some integer $m_2$ such that $(n_2 \alpha \bmod 1) = n_2 \alpha + m_2$.
We want to show that it's impossible to have $n_1 \alpha + m_1 = n_2 \alpha + m_2$. That's because if this holds, we can solve for $\alpha$ and get $\alpha = \frac{m_2 - m_1}{n_1 - n_2}$. But $\alpha$ was supposed to be irrational, contradiction. $\square$
Therefore $f$ is a bijection between $\mathbb Z$ and $A$. By definition, $|A| = |\mathbb Z|$: since $\mathbb Z$ is countably infinite, so is $A$.
