Problem 4 From Quiz like problems

Problem 4 states:

Suppose that $S$ is non-empty and bounded above and let $c\in\mathbb{R}$. Show that $S+c$ is non-empty and bounded above and that $\sup{(S+c)}=\sup{(S)} +c$.

Proof:

Let $S$ be non-empty and bounded above. Define
$$S +c = \{x+c:x\in{S}\}.$$
By the A.O.C, there is a least upper bound of $S$, namely $\sup(S)$.

Note that $\sup(S) \geq a$, for all $a\in S$. Thus, $\sup(S) + c \geq a + c$ for all $a\in S$ and $\sup(S) + c$ is an upper bound of $S + c$. By A.O.C, there is a least upper bound of $S+c$, namely $\sup(S+c)$.

Let $x$ be any upper bound of $S$ and notice $$x\geq \sup(S)\geq a \implies x+c \geq \sup(S) +c\geq a + c$$ Thus, $\sup(S) + c$ is the least upper bound for $S + c$.

Hence, $$\sup(S + c) = \sup(S) + c$$

End Proof
My thanks to Taylor...and looking forward to some feedback and critique!

My proof:

Since $S$ is bounded above and non-empty, by the Axiom of Completeness we may let $r = \sup(S)$.

Consider the following: $r \geq x, \; \forall \, x \in S \implies r+c \geq x+c, \; \forall \; (x+c) \in S+c$.

Suppose $q < r$. Thus, $\exists \, y \in S$ such that $y > q$.

This implies that if $q + c < r + c$, then $y + c > q + c$, where $(y+c) \in S+c$.

Therefore, $\sup(S+c) = r + c \implies \sup(S+c) = \sup(S) + c$. $\; \,\blacksquare$