Is the following argument correct?
Proposition. The set $T = \{1,1/2,1/3,1/4,1/5,\dots,1/n,\dots\}$ is not closed in $\mathbf{R}$.
Proof. Assume for the purpose of contradiction that the set $T$ is closed in $\mathbf{R}$, thus its complement $\mathbf{R}\backslash T = (-\infty,0)\cup\{0\}\cup\bigcup_{n=1}^{\infty}(\frac{1}{n+1},\frac{1}{n})\cup(1,\infty)$ is open. Then in particular for $x = 0$, there exists $a,b\in\mathbf{R}$ such that $a<b$ and $x\in(a,b)\subseteq\mathbf{R}\backslash T$, but then $x\in (a,\min\{b,1\})\subseteq\mathbf{R}\backslash T$.
Now the archimedian property of $\mathbf{R}$ implies that for some $n\in\mathbf{N}$ we have $n>\frac{1}{\min\{b,1\}}$, equivalently $\frac{1}{n}<\min\{b,1\}$, then since $a<x<\frac{1}{n}<\min\{b,1\}$ we have $\frac{1}{n}\in(a,\min\{b,1\})$ and thus $\frac{1}{n}\in\mathbf{R}\backslash T$, which is absurd considering that $T\cap\mathbf{R}\backslash T = \varnothing$.
$\blacksquare$