Following notation and terminology as in Terence Tao's "Analysis I", as it seems this question is taken from Exercise 6.1.8 (p. 132) where the reader is asked to prove Theorem 6.1.19 (p. 130).
We need to show that for any $\epsilon > 0$, the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^\infty$ is eventually $\epsilon$-close to $\max(x, y)$. Let $\epsilon > 0$ be arbitrary.
We know that we can find two natural numbers $N_1, N_2\geq m$ such that
$$
\begin{align*}
|a_n - x| \leq \epsilon \quad \text{and} \quad |b_n - y| \leq \epsilon
\end{align*}
$$
for all $n \geq \max(N_1, N_2)$. Let $N = \max(N_1, N_2)$.
We have two cases to consider: a) when $x = y$ and b) when $x \neq y$.
Assume first that $x = y$. In that case, $\max(x, y) = x = y$, so our task reduces to showing that the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $x$.
We know that for all $n \geq N$ we have both $|a_n - x| \leq \epsilon$ and $|b_n - x| \leq \epsilon$ (note the use of $x$ instead of $y$ under the assumption $x = y$). Hence, no matter whether $\max(a_n, b_n)$ is equal to $a_n$ or $b_n$ we are guaranteed that $|\max(a_n, b_n) - x| \leq \epsilon$ for all $n \geq N$. So the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $\max(x, y)$.
Now, for the second case assume that $x \neq y$. By the trichotomy of order on the real numbers we must have either $x < y$ or $x > y$. We really need to check both cases, but it turns out that the arguments used for one case can be tailored to fit with the other case. So we assume that $x > y$.
This assumption tells us that $\max(x, y) = x$. We therefore need to show that the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $x$. Again, we know that for all $n \geq N$
$$
\begin{align*}
|a_n - x| \leq \epsilon \quad \text{and} \quad |b_n - y| \leq \epsilon.
\end{align*}
$$
More specifically, we can set $\epsilon = \tfrac{x - y}{2}$ (note that since $x > y$ by assumption, this is a positive number, so $\epsilon > 0$.)
Using the fact (shown in Exercise 5.4.6) that
$$
\begin{align*}
|a_n - x| \leq \epsilon \implies x - \epsilon \leq a_n \leq x + \epsilon \\
|b_n - y| \leq \epsilon \implies y - \epsilon \leq b_n \leq y + \epsilon
\end{align*}
$$
we want to show that, at some point $a_n \geq b_n$, because then at some point $\max(a_n, b_n) = a_n$.
Using our epsilon from above, we have that $b_n \leq y + (x - y)/2 = (x + y)/2$. Similarly, we have that $x - (x-y)/2 = (x + y)/2 \leq a_n$. So,
$$
b_n \leq \frac{x+y}{2} \leq a_n
$$
for all $n \geq N$. Hence for all $n \geq N$ we have $|\max(a_n, b_n) - x| = |a_n - x| \leq \epsilon$. Consequently, the sequence $\left\{\max(a_n, b_n)\right\}_{n=m}^{\infty}$ is eventually $\epsilon$-close to $\max(x, y)$.
This concludes the proof.