Proving (directly) the equivalence between two definitions of a Fréchet or $T_1$ space.

Question

A space is Fréchet or $T_1$ if any of the following equivalent conditions hold:

1. Any two points of $X$ are separated i.e. there exist neighborhoods $U_x$ of $x$ and $U_y$ of $y$ such that $x\notin U_y$ and $y\notin U_x$.
2. Any finite subset of $X$ is closed.
3. Any singleton set $\{x\}\subseteq X$ is closed.
4. Every subset of $X$ is the intersection of all the open sets containing it.

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In proving that these conditions are equivalent, I have been able to show that $1\Rightarrow 2 \Rightarrow 3 \Rightarrow 4$, but I'm struggling in showing that $4\Rightarrow 1$. I was wondering if this can be done "directly" i.e. without cycling back to either of the conditions $2$ or $3$, since that would make the proof somewhat awkard.

An "indirect" proof can be constructed by showing that $1\Leftrightarrow 3$ (as done here), that $2\Leftrightarrow 3$ (which is easy) and that $2\Leftrightarrow 4$ (as done here).

Answer 1

Suppose 4) holds and pick $x,y \in A$ with $x \neq y$. Then $\{x\}$ is the intersection of all open sets containing $x$. Since $y \notin \{x\}$ it follows that $y$ cannot belong to every open set containing $x$. In other words, there exists an open set $U$ containing $x$ such that $y \notin U$. Similarly, there exists an open set $V$ containing $y$ such that $x \notin V$.

Answer 2

Asume condition 4 holds. Fix $x,y\in X$. $X\setminus\{x\}$ is the intersection of all open sets containing it, so it itself must be open (if it were not open, then the only open set containing it would be $X$). Similarly with $X\setminus\{y\}$. Then set $U_x = X\setminus\{y\}$ and $U_y= X\setminus\{x\}$.