A uniformly continuous map from a dense subspace into a complete metric space extends uniquely to a uniformly continuous map on the whole space

For every $x \in X$ and $n \in \mathbb{N}$ the set $U_n(x)$ is nonempty by [A1], so $S_n(x)$ is nonempty and $T_n(x)$ is a nonempty closed subset of $Y$.

The radii decrease, so $U_{n+1}(x) \subseteq U_n(x)$ and $S_{n+1}(x) \subseteq S_n(x)$; since $T_n(x)$ is a closed superset of $S_{n+1}(x)$, minimality of the closure gives $T_{n+1}(x) \subseteq T_n(x)$.

Fix a real $\varepsilon > 0$, let $\delta > 0$ be as in [A2] for $\varepsilon/3$, and let $N$ be a natural with $2/(N+1) < \delta$; note that $N$ depends on $\varepsilon$ alone and not on $x$.

Towards uniform continuity, let $\varepsilon > 0$ be real, let $\delta > 0$ be as in [A2] for $\varepsilon/3$, and put $\delta' := \delta/3 > 0$. Fix a natural $m$ with $1/(m+1) < \delta/3$.

For claim 2, let $h : X \to Y$ be continuous with $h(a) = f(a)$ for all $a \in A$, and let $x \in X$. Since $\overline{A} = X$ there is a sequence $(a_k)$ in $A$ with $a_k \to x$ in $X$.

Let $n \ge N$ and $a, a' \in U_n(x)$. Then $d_X(a,a') \le d_X(a,x) + d_X(x,a') < 2/(n+1) \le 2/(N+1) < \delta$, so $d_Y(f(a),f(a')) < \varepsilon/3$. Hence all pairwise distances in $S_n(x)$ are below $\varepsilon/3$, so $S_n(x)$ is bounded and $\operatorname{diam}(S_n(x)) \le \varepsilon/3$.

Let $n \ge N$, let $u,v \in T_n(x)$ and let $\eta > 0$ be real. The balls $B_Y(u,\eta)$ and $B_Y(v,\eta)$ meet $S_n(x)$, so there are $s,s' \in S_n(x)$ with $d_Y(u,s) < \eta$ and $d_Y(v,s') < \eta$, whence $d_Y(u,v) \le d_Y(u,s) + d_Y(s,s') + d_Y(s',v) < \varepsilon/3 + 2\eta$. As $\eta > 0$ was arbitrary, $d_Y(u,v) \le \varepsilon/3$: were $d_Y(u,v) > \varepsilon/3$, the value $\eta := (d_Y(u,v) - \varepsilon/3)/3$ would be positive and would give $d_Y(u,v) < d_Y(u,v)$.

So for $n \ge N$ the set $T_n(x)$ is nonempty, closed and bounded with $\operatorname{diam}(T_n(x)) \le \varepsilon/3 < \varepsilon$.

Apply steps 1.3 to 4.1 with $\varepsilon = 1$ to get a natural $N_1$ such that $T_n(x)$ is nonempty, closed and bounded for every $n \ge N_1$ and every $x \in X$. Then $\big(T_{N_1+j}(x)\big)_{j \in \mathbb{N}}$ is nested by step 1.2, and its diameters tend to $0$: given a real $\varepsilon > 0$, the $N$ of step 1.3 satisfies $\operatorname{diam}(T_{N_1+j}(x)) < \varepsilon$ for every $j \ge N$, since then $N_1 + j \ge N$.

By [L1] and [A3] the intersection $\bigcap_{j \in \mathbb{N}} T_{N_1+j}(x)$ has exactly one element; and because the family $(T_n(x))_n$ is nested this intersection equals $\bigcap_{n \in \mathbb{N}} T_n(x)$, a set defined without reference to $N_1$. Define $g(x)$ to be its unique element; this determines a function $g : X \to Y$, and no choice is made, since the value is unique.

$g$ extends $f$: for $a \in A$ and every $n$ we have $a \in U_n(a)$, so $f(a) \in S_n(a) \subseteq T_n(a)$; hence $f(a) \in \bigcap_n T_n(a)$, and by uniqueness $g(a) = f(a)$.

Let $x,x' \in X$ with $d_X(x,x') < \delta'$. Since $g(x) \in T_m(x) = \overline{S_m(x)}$, the ball $B_Y(g(x), \varepsilon/3)$ meets $S_m(x)$, so there is $a \in U_m(x)$ with $d_Y(g(x), f(a)) < \varepsilon/3$; likewise there is $a' \in U_m(x')$ with $d_Y(g(x'), f(a')) < \varepsilon/3$.

Then $d_X(a,a') \le d_X(a,x) + d_X(x,x') + d_X(x',a') < \delta/3 + \delta/3 + \delta/3 = \delta$, so $d_Y(f(a),f(a')) < \varepsilon/3$, and therefore $d_Y(g(x),g(x')) \le d_Y(g(x),f(a)) + d_Y(f(a),f(a')) + d_Y(f(a'),g(x')) < \varepsilon$.

The real $\delta'$ depended on $\varepsilon$ alone, so $g$ is uniformly continuous; together with step 7.1 this establishes claim 1.

The map $g$ is continuous, being uniformly continuous, so $g(a_k) \to g(x)$ and $h(a_k) \to h(x)$; but $g(a_k) = f(a_k) = h(a_k)$ for every $k$, so one sequence in $Y$ converges to both $g(x)$ and $h(x)$, whence $g(x) = h(x)$ by uniqueness of limits. As $x$ was arbitrary, $h = g$.

The map $g$ of step 6.1 is a uniformly continuous extension of $f$ and is the only continuous one, which is claims 1 and 2.