For the set $B=\left\{\left(t,\frac1t\right): t>0\right\}$ you argue that as $x$ goes to $\infty$, $1/x$ goes to $0$, but $0\notin B$, so $B$ isn’t closed. The problem with this argument is that $B$ isn’t a set of real numbers: it’s a set of ordered pairs of real numbers. The question isn’t whether $0\in B$ or not: we know that $0\not \in B$, because it isn’t even the kind of thing that can belong to $B$. The question is whether there is some point $(a,b)\notin B$ such that every open nbhd of $(a,b)$ contains a point of $B$.
Suppose, then, that the point $p=(a,b)\notin B$. If you make a rough sketch of $B$, you shouldn’t have any trouble convincing yourself that if $p$ is below and to the left of $B$, there are real numbers $c>a$ and $d>b$ such that $\{(x,y):x is an open nbhd of $p$ disjoint from $B$: it’s a lower lefthand ‘quadrant’ of the plane lying entirely below and to the left of $B$.
Similarly, if $p$ is above and to the right of $B$, there are real numbers $c and $d such that $\{(x,y):x>c\text{ and }y>d\}$ is an open nbhd of $p$ disjoint from $B$: it’s an upper righthand ‘quadrant’ of the plane lying entirely above and to the right of $B$.
Thus, every point of $\mathbb{R}^2\setminus B$ has an open nbhd disjoint from $B$, and $B$ is therefore closed.
If you actually had to prove that $B$ is closed, you’d have to explain how to find $c$ and $d$ for different values of $a$ and $b$, but if you’re simply trying to decide for yourself whether $B$ is closed, this sort of pictorial reasoning is just fine.