If $A$ has no max and $B$ is finite, then $\sup(A)=\sup(A\setminus B)$

Question

Let $A\subset \mathbb{R}$ be non-empty and bounded from above, and assume it does not have a maximum.
Let $B$ be a finite set of real numbers.

Prove: $\sup(A)=\sup(A\setminus B)$

Suppose: $\sup(A)\neq \sup(A\setminus B)$.
So there are $\sup(A)=t$, $\sup(A\setminus B)=s$ (w.l.o.g) $a< t\leq s- \epsilon< a-b \leq s$ but $B$ is finite so $\forall b\in B:B\in \mathbb{N,Z}$ therefore there is $a-b< s-\epsilon \leq s$, so that $s$ is not $\sup(A\setminus B)$, contradiction.

Is it valid proof? is there a straightforward proof?

Answer 1

Let $a=sup(A)$, $b=Sup(A\setminus B)$.

For all $x\in A\setminus B$ we have $x\leq b$.

Since $B$ has finite many elements, it has a maximum element $s$.

If $s\leq b$ then $a=b$.

If $b\leq s$ then $a=s$, since for all $x\in A \setminus B$: $x\leq b\leq s$ and for all $x\in B$: $x\leq s$. But $s$ is thus a maximum in $A$.