Proof of existence of a point in a closed set $C$ closest to a point not in $C$

Question

Note: The question was previously asked in Prove that there exists a nearest point in a closed set $A \subset \mathbf{R}^n$ to a point outside of $A$ answered with a wonderful simple proof.

This question is to clarify a similar but alternate proof of the same theorem in "The Mathematics of Nonlinear Programming" by Peressini, Sullivan, and Uhl.

Theorem: If $C$ is a closed (convex or not) subset of $\mathbb{R}^n$ and if $y \in \mathbb{R}^n$ does not belong to $C$, then there is a vector $x^* \in C$ that is closest to $y$, that is, $\forall x \in C$: $$\left\Vert y - x^* \right\Vert \leq \left\Vert y - x\right\Vert$$ The first half of the proof is stated:

Let $\alpha$ be the largest number such that $\forall x \in C, \ \alpha \leq \left\Vert y -x \right\Vert$. Then there is a sequence $\{x^{(k)}\}$ of elements of $C$ such that: $$\alpha = \lim_k \left\Vert y - x^{(k)} \right\Vert$$

Comment: $\alpha$ is just the infinium of the distance of $C$ and $y$

Question: How did we get to the existence of sequence $\{x^{(k)}\}$?

I understand $C$ is a closed set such that limit of a sequence in $C$ is also in $C$, but I don't think that's relevant to the distance function. I knew before that the distance function or $\left\Vert y - x \right\Vert$ for fixed $y$ is a continuous function, and I think that's the key, but the book has omitted this property, as in it's not mentioned anywhere and this book is more for undergraduates with little experience in Analysis.

Answer 1

This has nothing to do with $C$ being closed or any property of this set, this has just to do with the defining property of the infimum of a set which permit to deduce the following property.

Let $A\subset \mathbb{R}$ be a subset of the real line and $\alpha:=\inf\,A$ its infimum, then there is a sequence $\{a_n\}$ of elements of $A$ such that $\lim_{n\to\infty}\,a_n=\alpha$.

In order to prove this result, just note that as $\alpha$ is the infimum of $A$, for each $n\in\mathbb{N}$, there is some $a_n\in \left[\alpha,\alpha+\frac{1}{n}\right)$. Otherwise, $\alpha$ would not be the infimum as $\alpha+\frac{1}{2n}$ would be a lower bound for $A$. Then this $\{a_n\}$ is the sequence living in $A$ and converging to $\alpha$.

Once this is clear, the existence result for your sequence is just applying the above proposition to the set $\{||y-x||\,|\,x\in C\}$.