Conjecture
If $X$ is firt countable and if $f:X\rightarrow Y$ is a function then $f$ is not continuous at $x_0$ if this is an isolated point for $X$.
If $X$ is first countable then there exist a countable base $\mathscr{B}(x_0):=\{B_n\in\mathcal{U}(x_0):n\in\Bbb{N}\}$ for $x_0$ and so if this is an isolated point for $X$ there exist $n_0\in\Bbb{N}$ such that $B_{n_0}\cap X=\{x_0\}$. Now we suppose that the basic neighborhood of $\mathscr{B}(x_0)$ are such that $B_m\subseteq B_n$ for any $m\ge n$ so that if for any $n\in\Bbb{N}$ we pick a $x_n\in\ B_n$ then we make a sequcence $(x_n)_{n\in\Bbb{N}}$ converging to $x_0$ and additionally for any $n\ge n_0$ it must be $x_n=x_0$. Now we suppose that the function $f$ is continuous so that the sequence $f(x_n)$ converges to $f(x_0)$ and so if the statement if true I shall obtain a contradiction but unfortunately I can't do this so I ask to complete the proof if the statement it true and if not to take a counterexample and to show if the statement if true with additional (we can suppose that $Y$ is hausdorff or first countable) hypotheses because for example any real function if continuous in a isolated point. So could someone help me, please?
This is certainly not true as stated. Take $f$ to be a constant function.
In fact, for every function $f:X\to Y$, $f$ must be continuous at the isolated points because then for any open set $V\subseteq Y$ containing $f(x_0)$, there is an open set $U\subseteq f^{-1}(V)$ containing $x_0$, namely $U=\{x_0\}$.