Do I need induction to prove that this sequence is monotonic?

[CoffeeNerd999]

TL;DR Summary

If i'm proving $y_n = \sup\{x_j | j \geq n\}$ is decreasing, can this be done without invoking mathematical induction?

I think the initial assumptions would allow me to prove this without induction.

Suppose $(x_n)$ is a real sequence that is bounded above. Define $$ y_n = \sup\{x_j | j \geq n\}.$$

Let $n \in \mathbb{N}$. Then for all $j \in \mathbb{N}$ such that $j \geq n + 1 > n$
$$ x_{j} \leq y_n.$$ So, $y_n$ is an upper bounded of $(x_j)_{j=n+1}^\infty$. By definition, $y_{n+1}$ is the least upper bound of $(x_j)_{j=n+1}^\infty$, so
$$y_{n+1} \leq y_n.$$ Since $n$ was chosen arbitrarily, this proves $y_n$ is monotone decreasing.

Am I correct that the using the supremum's definition makes using the induction principle suprifilous for this result?

 

Your proof is correct and your observations as well. You don't need mathematical induction here. The following is the general situation that you applied: $$\emptyset \neq A \subseteq B \implies \sup A \leq \sup B$$

This is rather straightforward to prove (essentially the same proof you gave): If $\sup B = +\infty$, there is nothing to prove. So we may assume that $\sup B < +\infty$. Recall that $\sup B$ is by definition an upperbound for all elements in $B$, thus also for all elements of $A$ because $A \subseteq B$. By definition of supremum, we must have $\sup A \leq \sup B$ ($\sup A$ is the smallest upper bound for $A$ and $\sup B$ is an upper bound for $A$).

Since, $\{x_j: j \geq n\} \supseteq \{x_j: j \geq n+1\}$, this will allow you to conclude that $y_n \geq y_{n+1}$, which is what you want.