Consider a partially ordered set $(X,\leq)$. For finite subsets $A,B\subseteq X$, write $A\preceq B$ whenever there is an injection $f:A\to B$ such that $a\leq f(a)$ for every $a\in A$.
It is easily seen that the finite subsets of $X$ are now quasi-ordered by $\preceq$, that is, $\preceq$ is a reflexive and transitive relation.
I am wondering if the antisymmetry of $\leq$ now also extends to $\preceq$?
On
I think it does. Let $f:A\to B$ and $g:B\to A$ be injections such that $\forall a\in A\quad a\le f(a)$ and $\forall b\in B\quad b\le g(b).$ Let us prove that $A\subset B$ (by symmetry, $B\subset A$ will also hold).
Take some $a\in A$ and consider $$a\le f(a)\le g(f(a))\le f(g(f(a)))\le\dots.$$ Since $A$ is finite, there must exist two integers $0\le p<q=p+n$ such that $(g\circ f)^p(a)=(g\circ f)^q(a).$ Going backwards by injectivity of $f$ and $g$, this entails $a=(g\circ f)^n(a)$ hence $$a=f(a)=g(f(a))=f(g(f(a)))=\dots=(g\circ f)^n(a),$$ in particular $a=f(a)\in B,$ q.e.d.
On
The answer to your question is affirmative and can be established in a logically clear fashion along the following line of reasoning. Recall that an ordered set $(A, R)$ (where $R$ is an order relation on $A$) is called Artinian provided that any nonempty subset $\varnothing \neq X \subseteq A$ admits a minimal element (with respect to $R$, of course). Note that in particular finite ordered sets are always Artinian.
One first establishes the following fundamental
Lemma: Let $(A, R)$ be Artinian and $f \colon A \to A$ a surjection such that $x \leqslant_R f(x)$ for any $x \in A$ (let us agree to call such a map inflationary). Then necessarily $f=\mathbf{1}_A$ (the identity map of $A$).
Proof: we will use a method of reasoning preeminently featured in the theory of Artinian sets, namely the general form of induction, which in our case is to be formulated as follows: if for arbitrary $y \in A$ we have that $f(x)=x$ for any $x<_R y$, it follows that also $f(y)=y$. To show this is indeed the case for arbitrary $y \in A$, by the surjectivity of $f$ we gather the existence of $t \in A$ such $f(t)=y$; it follows immediately from the inflationary character of $f$ that $t \leqslant_R y$. Should it be the case that $t \neq y$ we would have automatically that $t <_R y$ and hence $f(t)=t \neq y$, by virtue of the (generalised) induction hypothesis. This leaves us with the only option $t=y$, entailing $f(y)=y$ and concluding the proof. $\Box$
This applies to the particular context of your problem in the following manner. Consider arbitrary ordered set $(A, R)$ and on the power-set $\mathscr{P}(A)$ define the binary relation $\mathscr{R}$ by: $$X\mathscr{R}Y \Leftrightarrow (\exists f)(f \in \mathrm{Hom}^{\mathrm{i}}_{\mathbf{Ens}}(X, Y)\ \wedge (\forall x)(x \in X \Rightarrow x\leqslant_R f(x))),$$
precisely as you have specified above. I have taken the liberty to use the notation $\mathrm{Hom}^{\mathrm{i}}_{\mathbf{Ens}}(M, N)$ to refer to the set of all injections from $M$ to $N$.
As you remark, $\mathscr{R}$ will in general only be a preorder on the powerset $\mathscr{P}(A)$, however its restriction to the collection of all subsets of $A$ which are Artinian with respect to $R$ becomes an order, satisfying the more stringent axiom of antisymmetry. Let us formalise this by introducing the notations:
With these preparations in place, we can claim that:
Proposition: $\left(\mathscr{Art}_R(A), \mathscr{R}_{|\mathscr{Art}_R(A)}\right)$ is an ordered set.
Proof: the claim amounts to saying that whenever $X \leqslant_{\mathscr{R}}Y$ and $Y \leqslant_{\mathscr{R}}X$ simultaneously hold it follows with necessity that $X=Y$, for arbitrary artinian subsets $X, Y \subseteq A$. We prove this directly, by first inferring the existence of injections $f \colon X \to Y$ respectively $g \colon Y \to X$ such that $x \leqslant_R f(x)$ for all $x \in X$ respectively $y \leqslant_R g(y)$ for all $y \in Y$, on the grounds of the sheer definition of $\mathscr{R}$.
It is clear that $g \circ f \colon X \to X$ is a bijection inflationary with respect to $R_{|X}$ and - since $(X, R_{|X})$ is Artinian by hypothesis - the above lemma applies with the conclusion that $g \circ f=\mathbf{1}_X$ (obviously, the analogous reasoning applies to the reverse composition $f \circ g=\mathbf{1}_Y$).
Considering the inflationary character of $f$ and $g$ - in this order of application, so to speak - we infer the relation $x \leqslant_R f(x) \leqslant_R g(f(x))=x$ for any $x \in X$, which by virtue of the antisymmetry of $R$ entails $f(x)=x$ for any $x \in X$ and hence $X \subseteq Y$; the reciprocal inclusion is of course established by considering the corresponding reasoning, where applies $g$ first and then $f$. Thus, we can conclude that $X=Y$. $\Box$
As mentioned in a previous paragraph, finite ordered sets are Artinian. If we agree to denote the collection of finite subsets of $A$ by $\mathscr{F}(A)$, we can thus formulate the succinct relation $\mathscr{F}(A) \subseteq \mathscr{Art}_R(A)$. As such, by transitivity of restrictions it is clear that $\mathscr{R_{|\mathscr{F}(A)}}$ is a restriction of $\mathscr{R}_{|\mathscr{Art}_R(A)}$ and as the latter is an order we conclude on grounds of heredity that so is the former.
Yes, it does. It may be easiest to reason by induction. Note that if $A \leq B$ and $B \leq A$ then $|A| = |B|$. Trivially it is true for $A, B$ empty. Now suppose that our statement is true for all subsets smaller than $A, B$.
Suppose that $A, B$ are such that $A \leq B$ and $B \leq A$ by your definition, witnessed by $f: A \to B$ and $g: B \to A$. Now take $a \in A$ and consider $a \leq f(a) \leq g(f(a)) \leq f(g(f(a))) \leq \cdots$. This is an infinite increasing sequence in the set $A \cup B$, and thus must eventually repeat, let's say starting at element $b$. Because $\leq$ is anti-symmetric, it follows that we find $b = f(b) = g(f(b))$. Then we can consider $A \setminus \{b\}$ and $B \setminus \{b\}$, and restricting $f, g$ to those sets still gives us that $A \setminus \{b\} \leq B \setminus \{b\}$ and vice versa. But by the induction hypothesis, it follows that these sets are equal. Thus they are also equal when $b$ is added back in, completing the proof by induction.