How is 0 a limit point of $\{1/n\}_{n=1}^{\infty}$?

Say $x = 0$ and you choose $\epsilon = 3$, and you follow along the proof: $$ (x-\epsilon, x+\epsilon) = \{-3,-2,-1,0,1,2,3\} $$ which certainly does contain a point $y = 1$ where $y \in \{1/n\}_{n=1}^{\infty}$, with $y$ being distinct from $x$.

You made a mistake of understanding the interval $(x-\epsilon, x+\epsilon)$ here. It is not equal to the set $\{-3,-2,-1,0,1,2,3\}$ but equal the set of real numbers $y$ such that $-3<y<3$. Note that the set $(x-\epsilon, x+\epsilon)$ has infinitely many elements! Moreover, it contains not only the point $1\in A:=\{1,\frac12,\frac13,\cdots\}$: it contains every element in $A$.

As far as I can tell, since $0 \notin A$, then $A\setminus\{x\}$ is equivalent to $\{1/n\}_{n=1}^{\infty}$, and the distance between $0$ and $A$ would be $1$: $$ d(0, [1, \frac 12, \frac 13, \frac 14 ...)) = 1 $$

What might I be missing here?

First of all, $[1, \frac 12, \frac 13, \frac 14 ...)$ is not a valid set notation. you could at most write $A=\{1,\frac12,\frac13,\frac14,\cdots\}$. In order to understand why $$ d(0,A)=0, $$ you need to read the definition of the notation $d(0,A)$ word by word slowly: $$ d(0,A)=\text{glb}\{|0-a|:a\in A\}. $$ Be very careful that "glb" means the "greatest lower bound". Now, without the "glb" symbol, look at the set $\{|0-a|:a\in A\}$. It happens to be exactly $A$: $$ \{|0-a|:a\in A\}=\{a:a\in A\}=A. $$ Now your question becomes:

why is the greatest lower bound of $A$ is $0$ not $1$?

Note that $1$ is not even a lower bound of the set $A$.