Let A be a nonempty set, such that $z\in A$ and $\sup(A) \not \in A$
(a) $\sup(A \setminus \{z\}) = \sup(A)$
(b) generalise this to $\sup(A \setminus \{z_1 z_2 z_3 \dots z_n\})=\sup(A)$
Let us denote $A$ with $z$ removed by $B$
For the first part I know that removing a point from $A$ can only really lower the supremum, because we might remove the right "end point", so the only possibility for $\sup(B)$ is that $\sup(B) \leq \sup(A)$, we can also see this via this formal argument:
Let $y \in A $, but $y \neq z$, then we have that $y \leq \sup (B)$, but certainly we have that $y\in B$ so now this is an upper bound for $B$, since $\sup(B)$ is the lowest of bounds for $B$, we have: $$ \sup(B) \leq \sup(A)$$
I fail so see the other direction, the problem is if we start out with an arbitrary element of $B$ is that we cannot say anything about $z$, at least I don't see this.
For (b) we simply solve this by induction on $n$, here is a rough sketch, I think this approach is fine, but correct me if I am wrong:
Our base case will hold by question (a), our inductive hypothesis generalises this for arbitrary $k \in \mathbb{N} $ to $A \setminus \{z_1,z_2,\dots,z_k\}$ and then for our inductive step we first consider $A \setminus \{z_1,z_2,\dots,z_k \}=C $ this is useful notation to make things shorter.
We can now write that: $\sup(C\setminus\{z_{k+1}\})=\sup(C)$ by question $(a)$,
We now use our inductive hypothesis ($\sup(C)=\sup(A)$) and state that: $$\sup(A\setminus\{z_1,z_2,\dots,z_k,z_{k+1}\})= \sup(C\setminus\{z_{k+1}\})=\sup(C)=\sup(A)$$ By the principle of mathematical induction the desired relation holds. $\square$
Suppose that $\sup(B)<\sup(A)$. Then, take $x\in A$ such that $\sup(B)<x<\sup(A)$ (We can do it by definition of $\sup$ and because $\sup(A)\notin A$). Then, we have that $x\notin B$ hence $x$ must be $z$. But it is impossible because it would imply that $z=\sup(A)$. Then, we have that $\sup(A)=\sup(B)$.