I wanted to prove that point continuity does not imply local continuity. I have already seen examples like the function $f(x) = x^{2}$ if $x$ is rational and 0 if irrational, which are nowhere continuous but at zero. However, I wanted to construct myself a general form of functions which prove the latter fact too. For that , I relied on any sequence $\{a_{n}\}$ converging to $f(a)$, with $a_{n}$ different from $f(a)$ for all $n > M$ and define $g(x)$ to equal $a_{n}$ whenever $|x-a| = 0.5^{n}$ and $f(a)$ otherwise. Then, I have proven two facts:
i) $f$ is continuous at $a$
ii) for every $r>0$, $f$ is discontinuous at certain points in the interval $(a-r,a+r)$.
To prove i) let $E>0$ be given. Then, focus on all $x$ such that $|x-a| < 0.5^{n}$ where $n =N(E)$ from the convergence of $\{a_{n}\}$ to $f(a)$. Now, $f(x)$ will equal $a_{m}$ for $m>n$ or $f(a)$ so $|f(x)-f(a)|<E$ either way. To prove ii) focus on the points $a-0.5^{n}$ and $a+0.5^{n}$ (call them $p$) where $n$ is a natural number large enough to have those points in the interval $(a-r,a+r)$ and such that $n>M$. Now, we aim to show that $f$ is not continuous at those points by showing that the limit is $f(a)$ instead (which is different from $a_{n}$ since $n>M$). For any $E>0$, see that if $0<|x-p|<0.5^{(n+1)}$ then, $|f(x)-f(a)| = |f(a)-f(a)| = 0 < E$. Therefore all of this shows that even though $f$ is continuous at $a$, no matter how small intervals around a are taken, $f$ is discontinuous somewhere. Please, let me know what you think and add anything you like.