Let $X$ and $Y$ be topological spaces (not necessarily assumed to be Hausdorff or to have any additional property) and let $f:X\to Y$ be a given function. Is it true that $f$ is continuous iff for every $x\in X$, and every net $\{x_i\}_i$ converging to $x$, one has that $f(x_i)\to f(x)$.
PS: I have searched for this specific question in MSE and, although I found several posts discussing it (this, this, and this) with varying degress of objectivity, and using various additional hypothesis (Hausdorff, completely regular, first countable) I do not believe it can be found in the exact terms above.
I therefore thought it would be nice to register it here.
I am also providing an answer which I hope is similar to the one in The Book $\ddot \smile$
The proof of the "only if" part is well known, so let us prove the "if" part.
Assuming by contradiction that $f$ is discontinuous at a point $x$ in $X$, we will build a net $\{x_i\}_{i\in I}$ converging to $x$, such that $\{f(x_i)\}_{i\in I}$ does not converge to $f(x)$. The set of indices for our net will be the set $N_x$ formed by all neighborhoods of $x$, and viewed as a directed set with order given by reverse inclusion, namely, $$ U\geq V \Leftrightarrow U\subseteq V. $$
Since $f$ is discontinuous at $x$, there exists a neighborhood $U$ of $f(x)$ such that $f(V)\not\subseteq U$, for all $V\in N_x$. Therefore, for any such $V$ we may choose some $x_V\in V$ such that $f(x_V)\not\in U$.
It is then evident that the net $\{x_V\}_{V\in N_x}$ converges to $x$, while $\{f(x_V)\}_{V\in N_x}$ does not converge to $f(x)$.
PS: The same argument above shows that $f$ is continuous at a given point $x_0$, if and only if, for every net $\{x_i\}_{i\in I}$, converging to $x_0$, one has that $\{f(x_i)\}_{i\in I}$ converges to $f(x_0)$.