Proof

By hypothesis, $(a_n)_{n\in\mathbb N}$ is a real sequence, $D_n:=\{a_k:~k\ge n\}$ are¹ subsets of its carrier set.
Let $\sup D_n$ be the supremum of extended real numbers for each $D_n$.
Case 1: $(a_n)_{n\in\mathbb N}$ is bounded above.
- Then $D_n$ are bounded above for all $n\in\mathbb N.$
- Note that: * $D_{n+1}\subset D_n,$ * and thus any upper bound of $D_{n}$ is also an upper bound of $D_{n+1}.$
- Also note that, by definition of supremum: * $\sup D_{n+1}$ is both, an upper bound of $D_{n+1}$ and less than or equal to every upper bound of $D_{n+1},$ * $\sup D_{n}$ is both, an upper bound of $D_{n}$ and less than or equal to every upper bound of $D_{n},$ * By first note, $\sup D_{n+1}\le \sup D_n.$
- It follows, that the sequence $(\sup D_n)_{n\in\mathbb N}$ is monotonically decreasing.
Case $(a_n)_{n\in\mathbb N}$ is unbounded.
- Then $\sup D_n=+\infty$ for all $n\in\mathbb N.$
- The order relation $\sup D_{n+1}\le \sup D_n$ is not defined in this case.
- But formally, we assert the sequence $(\sup D_n)_{n\in\mathbb N}$ is monotonically decreasing.

∎

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Forster Otto: "Analysis 1, Differential- und Integralrechnung einer Veränderlichen", Vieweg Studium, 1983

Note: $D_n$ can be visualized as $$\begin{array}{c|ccccc} n&D_n&&&&\\ \hline 1&a_1&a_2&a_3&a_4&\ldots\\ 2&&a_2&a_3&a_4&\ldots\\ 3&&&a_3&a_4&\ldots\\ 4&&&&a_4&\ldots\\ \vdots&&&&&\ddots\\ \end{array}$$ ↩