Trivial question about set difference: does $B\subseteq A$ imply $B\setminus\{x\}\subseteq A\setminus\{x\}$?

Suppose $y\in B\setminus\{x\}$. Then $y\in B$ and, since $B\subseteq A$, we have $y\in A$. Further, $y\in B\setminus\{x\}$ implies that $y\neq x$, and so $y\in A\setminus\{x\}$. Therefore, $B\setminus\{x\}\subseteq A\setminus\{x\}$.

If $b\in B\setminus \lbrace x\rbrace$, then $b\in B$ and $b\neq x$. Since $B$ is a subset of $A$, then we know $b\in A$. So $b\in A$ and $b\neq x$, by definition it follows that $b\in A\setminus \lbrace x\rbrace$.

The statement $B\subseteq A$ means every member of $B$ is a member of $A$. The question is then whether every member of $B\smallsetminus\{x\}$ is a member of $A\smallsetminus\{x\}.$

Suppose $y$ is a member of $B\smallsetminus\{x\}.$ Then $y$ is a member of $B$ and $y\ne x.$ Since $y$ is a member of $B$ and every member of $B$ is a member of $A$, it follows that $y$ is a member of $A.$ And it is still true that $y\ne x.$

Thus every member of $B\smallsetminus\{x\}$ is a member of $A\smallsetminus\{x\}.$