Question on proving existence in Rudin exercise 1.5

This exercise from Rudin reads:

Let $A$ be nonempty set of real numbers which is bounded below. Let $-A$ be the set of all real numbers $-x$, where $x \in A$. Prove that $$\inf A = -\sup(-A).$$

I don't have a question on solving the problem, which I believe I know how to do. My only question is on whether, for full rigor, one must show the existence of $\inf A$ and $-\sup(-A)$.

After much thought, my conclusion is as follows. The proof proceeds by arguing that because $A$ is bounded below, $-A$ is bounded above (and nonempty since $A$ is nonempty), so it admits a supremum. That establishes the existence of $\sup(-A)$. But I don't think I need to argue that $\inf A$ exists, even though it's a simple statement of the greatest-lower-bound property, which is a theorem in Rudin. I think I can simply show that $-\sup(-A)$ satisfies the properties of the infimum, namely that it is a lower bound of $A$ and the greatest such lower bound.

Is this correct, or should I state outright, "$\inf A$ exists by the greatest-lower-bound property, and now we show that $-\sup(-A)$ is the desired infimum? I've seen multiple proofs of this fact proceed differently, and I want to be as rigorous as possible.

In definition 1.8 of Rudin, it is remarked that there is at most one least upper bound of a nonempty subset $E$ of $S$. Of course, the same is true when "least upper bound" is replaced with "greatest lower bound." Therefore, if the infimum of a nonempty subset $A$ of $\mathbb{R}$ exists, it must be unique (i.e. the only number in $\mathbb{R}$ satisfying the properties of the infimum).

So, if you are comfortable with this fact, then once you have shown $- \text{sup} \; A$ satisfies the properties of the infimum, by uniqueness you know that $- \text{sup} \; A$ must actually be the infimum. Therefore, you have $$\text{inf} \; A = - \text{sup} \; A$$ as desired, with no need to separately prove that $\text{inf} \; A$ exists.

However, if you haven't worked through the details of this remark, it is useful to do so.

Let's suppose $E$ is a subset of an ordered set $S,$ and that $E$ has an infimum. Further suppose that $e_{1}$ and $e_{2}$ are two infimums of $E$. By definition, $e_{1}$ and $e_{2}$ are both lower bounds of $E$. Then, since $e_{1}$ is an infimum of $E$, it is a greatest lower bound. Since $e_{2}$ is also a lower bound we have $$e_{2} \leq e_{1}.$$ However, since $e_{2}$ is also an infimum of $E$, and $e_{1}$ is a lower bound, by the same reasoning we conclude that $$e_{1} \leq e_{2}.$$

So, we must have $e_{1} = e_{2},$ and thus if an infimum of a nonempty subset $E$ of an ordered set $S$ exists, it is unique.