Let be $(E, d)$ a metric space. Let be $x \in E$ and $A \subset E$.
We define $d(x, A) = \inf\limits_{y \in A} d(x, y)$.
Let be $(A_n) \in E^{\mathbb{N}}$ so that $A_{n + 1} \subset A_n$ and $A = \bigcup\limits_{n \in \mathbb{N}} A_n$.
Then, we can show that $(d(x, A_n))_{n \in \mathbb{N}} \in \left(\mathbb{R}_{+}^{*}\right)^{\mathbb{N}}$ is a decreasing sequence, which converges to $d(x, A)$ by considering the property of the infinimum (i.e. $\forall \varepsilon > 0, \exists y \in A, 0 \leq d(x, y) \leq d(x, A) + \varepsilon$).
But as, $d(\cdot, A) : x \mapsto d(x, A)$ is a $1$-lipschitz map, I believe there could be a faster way to prove the previous statement.
Notice that for any $n$, $$d(x,A_n) ≥ d(x,A)$$ because for arbitrary functions $f$, if $U\subset V$ then $\inf_{v\in U} f(v) ≥ \inf_{v\in V}f(v)$. Similarly, $(d(x,A_n))_{n≥1}$ is a decreasing sequence.
To conclude that $d(x,A_n)$ decreases to $d(x,A)$, we show that for any $\epsilon>0$, there is $n_0\in\mathbb N$ such that for every $n \geq n_0$, $d(x,A_n) ≤ d(x,A) + \epsilon$, which implies the result.
To this end, let $\epsilon>0$ be fixed. The fact that $d(x,A)$ is an infimum means that there exists $y=y(\epsilon)\in A$ such that $$ d(x,y) ≤ d(x,A) + \epsilon$$ Since $A = \bigcup_{k=1}^\infty A_k$, there exists some $n_0=n_0(\epsilon)$ such that $y\in A_{n_0}$, and the sets $A_k$ are increasing so $y\in A_n $ for every $n≥n_0$. Thus for these $n$, $d(x,A_n) ≤ d(x,y)$, so that $$ d(x,A_n) ≤ d(x,A) + \epsilon$$ as needed.