Every convergent sequence is bounded: what's wrong with this counterexample?

The result is saying that any convergence sequence in real numbers is bounded. The sequence that you have constructed is not a sequence in real numbers, it is a sequence in extended real numbers if you take the convention that $1/0=\infty$.

A real sequence is nothing but a function $$f:\mathbb{N}\longrightarrow \mathbb{R}$$ One often writes for example $x_n$ instead of $f(n)$ etc.

So, your "sequence" is not defined on $\mathbb{N}$ but on $\mathbb{N}\setminus\{a\}$.

But nicely enough, if you remove the "undefined member" by setting $x_a := r$ to an arbitrary real number $r$, you get a convergent sequence which is bounded.

Your sequence $\{ 1/(n-a)\}_1^\infty $ is not defined at $n=a$ if $a$ is a positive integer.

Thus it is not a sequence at all.

for example $$\{ 1/(n-5)\} _1^\infty = \{ -1/4,-1/3,-1/2, -1, ?, 1,...\}$$