What condition(s) on $X$ and $A$ can ensure the existence of an element $a\in A$ such that $d(x_0, a)=d(x_0, A) $?

Question

Let $(X, d) $ be metric space and $A\subset X$ and $x_0\in X$

Define $d(x_0, A) =\inf\{d(x_0, a) :a\in A\}$

What condition(s) on $X$ and $A$ can ensure the existence of an element $a\in A$ such that $d(x_0, a)=d(x_0, A) $ ?

Can we take care of uniqueness as well?

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Motivation:

Suppose $X$ is a Hilbert space and $A\subset X$ non empty closed convex set then for all $x_0\in X$ $\exists! a\in A$ such that $d(x_0,a) =d(x_0, A) $

$X$ is reflexive and strictly convex iff $ A\subset X$ non empty closed convex set in $X$ is a Chebyshev set.

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Particular cases:

Suppose $(X,d)$ be any metric spaces $A\subset X$ non empty compact set then $\forall x_0\in X, \exists a\in A$ such that $d(x_0, a) =d(x_0, A) $

Sketch:

$d_{x_0}:X\to\Bbb{R}$ is continuous and $A\subset X$ compact implies $\inf\{d(x_0, a) :a\in A\}$ is attained in A.

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Going deep into the proof:

$d(x_0, A)=\inf\{d(x_0, a) :a\in A\}$ implies $\exists (a_n) \in A$ such that $d(x_0, a_n) \to d(x_0, A) $

$\lim _{n\to\infty}d(x_0, a_n)= d(x_0, A) $

Implies $ d(x_0,\lim_{n\to\infty} a_n)= d(x_0, A) $

We only need to make sure that $(a_n)\subset A$ converges to $a\in A$ for some $a\in A$

$$\text{OR}$$

$(a_n) $ has a convergent subsequence in $A$

Further treatment:

$(X, d) $ complete and $A\subset X$ non empty closed totally bounded set.

But no improvement at all! (Complete and totally boundedness $\iff$ compact.)