[proofplan] We prove the contrapositive-style contradiction: assume that the limiting order fails, so $x>y$. The positive gap $x-y$ lets us choose an error tolerance small enough that all sufficiently late $x_n$ remain above the midpoint of the gap and all sufficiently late $y_n$ remain below it. Taking one index beyond the convergence thresholds and the eventual-order threshold then gives $x_n>y_n$, contradicting the hypothesis $x_n\leq y_n$. [/proofplan]

[step:Assume a positive limiting gap and choose a common tolerance]Assume, for contradiction, that $x>y$. Define the positive real number \begin{align*} \varepsilon := \frac{x-y}{3}. \end{align*} Since $x-y>0$, we have $\varepsilon>0$.[/step]

[guided]We argue by contradiction because the hypothesis gives inequalities for the approximating sequences, while the negation of the desired conclusion gives a strict positive gap between the limits. Suppose that $x>y$. Then the distance between the two limits is the positive number $x-y$. We choose \begin{align*} \varepsilon := \frac{x-y}{3}. \end{align*} This choice is positive because $x-y>0$. The particular factor $1/3$ is not important; what matters is that the two convergence errors together are smaller than the gap $x-y$. With this choice, the total possible error $2\varepsilon$ satisfies \begin{align*} 2\varepsilon = \frac{2(x-y)}{3} < x-y. \end{align*} This strict inequality is what will force a late term of $(x_n)$ to lie above the corresponding late term of $(y_n)$.[/guided]

[step:Use convergence to control both sequences beyond fixed indices] Because $x_n\to x$, the definition of convergence of a real sequence applied with the tolerance $\varepsilon>0$ gives an index $N_x\in\mathbb{N}$ such that \begin{align*} |x_n-x|<\varepsilon \end{align*} for every $n\geq N_x$. Hence, for every $n\geq N_x$, \begin{align*} x_n>x-\varepsilon. \end{align*} Because $y_n\to y$, the same definition applied with the same tolerance $\varepsilon>0$ gives an index $N_y\in\mathbb{N}$ such that \begin{align*} |y_n-y|<\varepsilon \end{align*} for every $n\geq N_y$. Hence, for every $n\geq N_y$, \begin{align*} y_n<y+\varepsilon. \end{align*} [/step]

[step:Choose one index where convergence and eventual order all apply]Define \begin{align*} M := \max\{N,N_x,N_y\}. \end{align*} Then $M\in\mathbb{N}$ and $M\geq N$, $M\geq N_x$, and $M\geq N_y$. Since $M\geq N_x$ and $M\geq N_y$, the estimates from convergence give \begin{align*} x_M>x-\varepsilon \end{align*} and \begin{align*} y_M<y+\varepsilon. \end{align*} By the definition of $\varepsilon$, \begin{align*} x-\varepsilon = x-\frac{x-y}{3} = \frac{2x+y}{3} \end{align*} and \begin{align*} y+\varepsilon = y+\frac{x-y}{3} = \frac{x+2y}{3}. \end{align*} Since $x>y$, \begin{align*} \frac{2x+y}{3}>\frac{x+2y}{3}. \end{align*} Therefore, \begin{align*} x_M>x-\varepsilon>y+\varepsilon>y_M. \end{align*} Thus $x_M>y_M$.[/step]

[guided]We now need a single index for which all three facts are simultaneously available: the eventual inequality $x_n\leq y_n$, the closeness of $x_n$ to $x$, and the closeness of $y_n$ to $y$. Define \begin{align*} M := \max\{N,N_x,N_y\}. \end{align*} This maximum is a natural number because it is the maximum of finitely many natural numbers. By construction, $M$ is beyond every threshold. Since $M\geq N_x$, convergence of $(x_n)$ gives \begin{align*} |x_M-x|<\varepsilon. \end{align*} From $|x_M-x|<\varepsilon$ we get $x_M>x-\varepsilon$. Since $M\geq N_y$, convergence of $(y_n)$ gives \begin{align*} |y_M-y|<\varepsilon. \end{align*} From $|y_M-y|<\varepsilon$ we get $y_M<y+\varepsilon$. It remains to compare the two bounds $x-\varepsilon$ and $y+\varepsilon$. Using $\varepsilon=(x-y)/3$, we compute \begin{align*} x-\varepsilon = x-\frac{x-y}{3} = \frac{2x+y}{3}. \end{align*} Similarly, \begin{align*} y+\varepsilon = y+\frac{x-y}{3} = \frac{x+2y}{3}. \end{align*} Because $x>y$, subtracting the second displayed quantity from the first gives \begin{align*} \frac{2x+y}{3}-\frac{x+2y}{3}=\frac{x-y}{3}>0. \end{align*} Hence $x-\varepsilon>y+\varepsilon$. Combining the inequalities gives \begin{align*} x_M>x-\varepsilon>y+\varepsilon>y_M. \end{align*} Thus $x_M>y_M$.[/guided]

[step:Contradict the eventual order and conclude the limiting order] Since $M\geq N$, the hypothesis gives $x_M\leq y_M$. This contradicts the conclusion $x_M>y_M$ obtained above. Therefore the assumption $x>y$ is false. Since real numbers are linearly ordered, not having $x>y$ is equivalent to $x\leq y$. Hence $x\leq y$, as required. [/step]