Why from $xThis trick is used by Rudin in his "Functional Analysis" to prove a certain inequality:

In essence, he shows that, given some numbers $x,y,t$, if $x<t$, then necessarily $y\leq t$ (why this is true I understand) and that, because this works for every $t$ (that fulfills the inequality $x<t$), from this it automatically follows that $y\leq x$. The final implication I don't understand, why exactly does this
$$((x<t)\Rightarrow(y\leq t))\implies(y\leq x)$$
work?
I feel like an idiot for not understanding this, it seems like it should be trivial.

By way of contradiction, let
$$x< t \implies y\leq t$$
but $y>x$. Then let $t = \frac{x+y}2$ and observe how $x=\frac{x+x}2 <\frac{x+y}2 = t$ but $y=\frac{y+y}2 > \frac{x+y}2 = t$ so the implication fails. Thus you must have $y\leq x$.
Edit:
This may feel dodgy, so if you want to think more like someone wanting to understand analysis, you could think about how this holds for all $t$ such $x< t \leq y$. Thus you cannot have any $t$ such $x<t\leq y$, so you cannot have $y$ to be greater than $x$.
A convenient choice is usually their average, of course. And thinking BWOC is often a decent way to cheese analysis.
