The problem to solve is:
"Let $(X,d)$ be an non compact metric space then prove that there exists a continuous function $f$ such that $f:X \to \mathbf{R}$ is continuous but not bounded."
The proof I found on MathstackExchange is given as below,
"Here is an explicit and elementary construction, using only the fact that a metric space is compact iff every sequence has an accumulation point. Let $(x_n)$ be a sequence in $X$ with no accumulation point, and define $g(x)=\inf_n d(x,x_n)+1/n$ and $f(x)=1/g(x)$. It is easy to see that $g(x)=0$ iff $x$ is an accumulation point of $(x_n)$, so $g$ is never $0$ and $f$ is defined on all of $X$. Moreover, $f(x_n)\geq n$, so $f$ is unbounded. Finally, it is straightforward to check that $g$ is continuous, and hence so is $f$."
My question is what is the g defined here.. Is it well defined first? it is said that $g(x)=\inf_n d(x,x_n)+1/n$ what is the value of $\frac{1}{n}$ for a specific x, how does it varies, if it is $g(x)=\inf_n (d(x,x_n)+1/n)$ then $g(x_n)$ will be $0$ and so f will not be defined on those points.. so it is clearly not $g(x)=\inf_n (d(x,x_n)+1/n)$.. Kindly help..
Your error is suggesting that $$ g(x_m) = \inf_n \left( d(x_n,x_m) + \frac{1}{n} \right) = 0 $$ necessarily is true. While it is true that $$ d(x_m,x_m) + \frac{1}{m} = \frac 1 m $$ this does not mean $g(x_m) = 0$. (What if $d(x,x_n)$ is very small for very large $n$, but always nonetheless bounded away from zero?)
More succinctly, $$ \inf_n \left( d(x_n,x_m) + \frac{1}{n} \right) \ne \inf_n \left( d(x_n,x_n) + \frac{1}{n} \right) = \inf_n \frac 1 n $$
Note that slight rephrasing in what you suggest is true and what I write above in my first lines as well, and why I used $x_m$. The infimum runs over all possible values of $n$, so you confound the notation a bit by trying to include $x_n$ as an argument of $g$. Presumably what you would like to do - plug in just any member of the sequence - is reflected above, and hopefully you can see why this claim might be an issue.