If the inclusion $X\setminus B\subseteq X\setminus A$ holds then does the inclusion $A\subseteq B$ even when $A$ and $B$ are not contained into $X$?

Question

Let's we assume the inclusion $$ A\subseteq B $$ holds. So if $X\setminus B$ is empty then surely it is contained into $X\setminus A$ for any $X$; however if $x$ is in $X\setminus B$ then it is not in $A$ since otherwise it was in $B$: so we conclude that for any $X$ the inclusion $$ X\setminus B\subseteq X\setminus A $$ holds.

Now let's we assume the inclusion $$ X\setminus B\subseteq X\setminus A $$ for some $X$: so by what proved above we conclude that the inclusion $$ A\cap X=X\setminus(X\setminus A)\subseteq X\setminus(X\setminus B)=B\cap X $$ holds: so if $X$ do not contains $A$ and $B$ then it seems possible that $A$ is not contained into $B$ so that I thought to put a specific question where I ask for a counterexample, provided this exists. So could someone help me, please?

Answer 1

Suppose $X=\{0\}$, $A=\{1\}$, and $B=\{2\}$.