I was trying to prove the following (part (g) of the Theorem):
Theorem 6.1.19 (Limit Laws). Let $(a_n)_{n=m}^{\infty}$ and $(b_n)_{n=m}^{\infty}$ be convergent sequences of real numbers, andlet $x,y$ be the real numbers $x := \lim_{n \to \infty}a_n$ and $y :=\lim_{n \to \infty}b_n$.
(g) The sequence $(\text{max}(a_n,b_n))_{n=m}^\infty$ converges to$\text{max}(x,y)$; in other words,
$$ \lim_{n \to \infty} \text{max}(a_n,b_n)=\text{max}(\lim_{n \to\infty}a_n,\lim_{n \to \infty}b_n) $$
I have written out my attempt below, but the proof is incomplete. I have laid out two questions within the writing below that I'm hoping someone could help me with. I am aware there are solution to this already online, but I wanted to see if my attempt is worth fixing if possible.
Proof attempt.
By the hypothesis of the theorem, we have that for every $\varepsilon > 0$, there exist positive integers $N_1,N_2 \geq m$ such that
$$|a_n-x| \leq \varepsilon, \ \text{for all} \ n \geq N_1$$
and
$$|b_n-y| \leq \varepsilon, \ \text{for all} \ n \geq N_2$$
Assume, without loss of generality, that $x \leq y$, so that $\text{max}(x,y)=y.$
We want to show that for every $\varepsilon > 0$, there exists a positive integer $N \geq m$ such that $|\text{max}(a_n,b_n) - y| \leq \varepsilon$, for all $n \geq N$. Well, we have that
\begin{align*}|\text{max}(a_n,b_n) - y| &= |\text{max}(a_n,b_n)-b_n+b_n-y| \\&\leq |\text{max}(a_n,b_n) - b_n| + |b_n - y| \\\end{align*}
I want to bound $|\text{max}(a_n,b_n) - b_n|$. Well, for $n\geq m$, we have two cases depending on $n$. We could have $\text{max}(a_n,b_n)=b_n$ or $\text{max}(a_n,b_n)=a_n$. In the former case, we'd have $|\text{max}(a_n,b_n) - b_n|=|b_n-b_n|=0$. In the latter case, we'd have $|\text{max}(a_n,b_n) - b_n|=|a_n-b_n|\geq0$. Thus, regardless which case it is, we'll always have $|\text{max}(a_n,b_n) - b_n| \leq|a_n-b_n|$.
Question 1. Is my reasoning in the above paragraph valid?
Continuing where we left off, we then have
\begin{align*}|\text{max}(a_n,b_n) - y| &= |\text{max}(a_n,b_n)-b_n+b_n-y| \\&\leq |\text{max}(a_n,b_n) - b_n| + |b_n - y| \\&\leq |a_n-b_n| + |b_n-y| \\&= |a_n-x + x-b_n| + |b_n-y| \\&\leq |a_n-x| + |x-b_n| + |b_n-y| \\&= |a_n-x| + |b_n-y| + |b_n-y+y-x| \\&\leq |a_n-x| + |b_n-y| + |b_n-y| + |y-x| \\&\leq |a_n-x| + |b_n-y| + |b_n-y| + |y-x| + 1\end{align*}
With $\varepsilon > 0$ and $|y-x| + 1 > 0$, we can use the Archimedean Property to find a positive integer $M$ such that $|y-x| + 1 < M \varepsilon$. This implies that
\begin{align*}|\text{max}(a_n,b_n) - y| & \leq |a_n-x| + |b_n-y| + |b_n-y| + M \varepsilon\end{align*}
Question 2. At this point, I got stuck. I have been trying to find bounds for $|a_n-x|$ and $|b_n-y|$ such that we still have $\varepsilon > 0$ and that the right side of the inequality is either eqaul to $\varepsilon$ or less than it. But no matter what I thought of, nothing worked. Is this a lost cause? Should I think of a different strategy?
