Say a space $X$ is totally retracting if every nonempty subset of $X$ is a retract. Then I claim that $X$ is totally retracting iff its $T_0$ quotient is a disjoint union of spaces $X_i$ such that each $X_i$ is isomorphic to a finite chain, $\mathbb{N}$, $-\mathbb{N}$, or $\mathbb{Z}$ with the Alexandrov topology.

First, it is easy to see that $X$ is totally retracting iff its $T_0$ quotient is, so we may assume $X$ is $T_0$. Also, a disjoint union of totally retracting spaces is totally retracting, so the "if" direction is easy (see Emil's answer for the case of $\mathbb{N}$, and the others are similar).

Thus let $X$ be a totally retracting $T_0$ space; we want to write it as a disjoint union of simple chains with the Alexandrov topology. Let $\leq$ denote the specialization order on $X$ and let $\{X_i\}$ be the connected components of $X$ with respect to $\leq$. If $A\subseteq X$ contains one point from each $X_i$, then $A$ is $T_1$. Any subset of a totally retracting space is totally retracting, so it follows that $A$ is discrete (see Francois's comment). But any continuous map must preserve the specialization order, so it follows that if $r:X\to A$ is a retraction, the inverse images of any point of $A$ is the unique $X_i$ that contains it. Since $A$ is discrete, it follows that $X$ is the disjoint union of the $X_i$.

We thus may assume that $X$ is connected with respect to $\leq$, and we wish to show that $X$ is isomorphic to a finite chain, $\mathbb{N}$, $-\mathbb{N}$, or $\mathbb{Z}$ with the Alexandrov topology. First, since $X$ is connected, every subset of $X$ must be connected, and so $\leq$ must be a total order. Now suppose there is an infinite increaseing sequence $x_0<x_1<\dots$ in $X$ and an element $y\in X$ such that $x_n<y$ for all $n$. Let $A=\{x_0,x_1\dots\}\cup\{y\}$ as a subspace of $X$. Then $\{x_0,x_1\dots\}$ cannot be a retract of $A$ (there's nowhere for $y$ to go), so $A$ is not totally retracting. This is a contradiction, so there can be no infinite increasing sequence in $X$ that is bounded above. By a similar argument, there can be no infinite decreasing sequence that is bounded below.

It follows that for any $x<y$ in $X$ there are only finitely many $z$ such that $x<z<y$. From this it follows easily that as an ordered set, $X$ is isomorphic to either a finite chain, $\mathbb{N}$, $\mathbb{-N}$, or $\mathbb{Z}$. It remains to be shown that $X$ has the Alexandrov topology. To prove this, we must show that for any $x\in X$, $U_x=\{y:y\geq x\}$ is open. Let $x-1$ be the predecessor of $x$ with respect to the ordering (if $x$ is the least element, $U_x$ is all of $X$). Since $x\not\leq x-1$, there exists an open set $U$ that contains $x$ but does not contain $x-1$. It is now easy to see that $U=U_x$.

Say a space $X$ is totally retracting if every nonempty subset of $X$ is a retract. Then I claim that $X$ is totally retracting iff its $T_0$ quotient is a disjoint union of spaces $X_i$ such that each $X_i$ is isomorphic to a finite chain, $\mathbb{N}$, $-\mathbb{N}$, or $\mathbb{Z}$ with the Alexandrov topology.

First, it is easy to see that $X$ is totally retracting iff its $T_0$ quotient is, so we may assume $X$ is $T_0$. Also, a disjoint union of totally retracting spaces is totally retracting, so the "if" direction is easy (see Emil's answer for the case of $\mathbb{N}$, and the others are similar).

Thus let $X$ be a totally retracting $T_0$ space; we want to write it as a disjoint union of simple chains with the Alexandrov topology. Let $\leq$ denote the specialization order on $X$ and let $\{X_i\}$ be the connected components of $X$ with respect to $\leq$. If $A\subseteq X$ contains one point from each $X_i$, then $A$ is $T_1$. Any subset of a totally retracting space is totally retracting, so it follows that $A$ is discrete (see Francois's comment). But any continuous map must preserve the specialization order, so it follows that if $r:X\to A$ is a retraction, the preimage of any point of $A$ is the unique $X_i$ that contains it. Since $A$ is discrete, it follows that $X$ is the disjoint union of the $X_i$.

We thus may assume that $X$ is connected with respect to $\leq$, and we wish to show that $X$ is isomorphic to a finite chain, $\mathbb{N}$, $-\mathbb{N}$, or $\mathbb{Z}$ with the Alexandrov topology. First, since $X$ is connected, every subset of $X$ must be connected, and so $\leq$ must be a total order. Now suppose there is an infinite increaseing sequence $x_0<x_1<\dots$ in $X$ and an element $y\in X$ such that $x_n<y$ for all $n$. Let $A=\{x_0,x_1\dots\}\cup\{y\}$ as a subspace of $X$. Then $\{x_0,x_1\dots\}$ cannot be a retract of $A$ (there's nowhere for $y$ to go), so $A$ is not totally retracting. This is a contradiction, so there can be no infinite increasing sequence in $X$ that is bounded above. By a similar argument, there can be no infinite decreasing sequence that is bounded below.

It follows that for any $x<y$ in $X$ there are only finitely many $z$ such that $x<z<y$. From this it follows easily that as an ordered set, $X$ is isomorphic to either a finite chain, $\mathbb{N}$, $\mathbb{-N}$, or $\mathbb{Z}$. It remains to be shown that $X$ has the Alexandrov topology. To prove this, we must show that for any $x\in X$, $U_x=\{y:y\geq x\}$ is open. Let $x-1$ be the predecessor of $x$ with respect to the ordering (if $x$ is the least element, $U_x$ is trivially open). Since $x\not\leq x-1$, there exists an open set $U$ that contains $x$ but does not contain $x-1$. It is now easy to see that $U=U_x$.