Let $T = (X, \mathcal{T})$ be a topological space.
Let $T$ be a Fréchet space. The implication holds vacuously in the case that $X$ is empty, so suppose that $X$ is nonempty and let $x \in X$. If $X$ is a singleton containing only $x$, then $X \setminus \{ x \}$ is empty and therefore, by definition, open in $T$.
We may thus assume that $X$ has at least two elements. Then $X \setminus \{ x \}$ is not empty and there is some $y \in X \setminus \{ x \}$. Since $T$ is a Fréchet space, there is an open neighbourhood $U$ of $y$ such that $x \not\in U$. Hence $U \subseteq X \setminus \{ x \}$. Since $y \in X \setminus \{ x \}$ was arbitrary, result
R458: Open set iff contains an open neighbourhood at every point implies that $\{ x \}$ is open in $T$. Since $x \in X$ was arbitrary, this establishes the implication.
Conversely, suppose that singletons are closed in $T$. Again, if $X$ has less than two elements then the implication holds vacuously, so we may assume $|X| \geq 2$. Let $x, y \in X$ such that $x$ does not equal $y$. Since singletons are closed in $T$, then both $X \setminus \{ x \}$ and $X \setminus \{ y \}$ are open in $T$. Furthermore, $x$ belongs to $X \setminus \{ y \}$, but not to $X \setminus \{ y \}$. Similarly, $y$ belongs to $X \setminus \{ x \}$ but not to $X \setminus \{ y \}$. This confirms the Fréchet condition, and the proof is finished. $\square$
