A pointwise convergent subsequence of a uniformly Lipschitz family into a compact metric space converges uniformly on compact subintervals

(Deterministic diagonal extraction after countable choice.) Enumerate $D=\{q_1,q_2,\dots \}$ [L1]. Compactness gives a finite cover of $B$ by balls of radius $1\mathrm{/}n$ for each $n\ge 1$; [L3] chooses one ordered finite list of centers for each $n$, from this family fixed in advance. Set $A_0=\mathbb{N}$. At stage $n$, classify the indices $m\in A_{n-1}$ by the tuple of least ball indices containing ${\sigma }_m(q_1),\dots ,{\sigma }_m(q_n)$ in the chosen radius-$1\mathrm{/}n$ cover. There are finitely many classes, so at least one is infinite; let $A_n$ be the infinite class with lexicographically least tuple. This defines nested infinite sets by deterministic recursion, without dependent choice. Put $m_0=0$ and let $m_n$ be the least member of $A_n$ greater than $m_{n-1}$. For fixed $i$ and $n\ge i$, every $m_j$ with $j\ge n$ lies in $A_n$, so the values ${\sigma }_{m_j}(q_i)$ lie in one ball of radius $1\mathrm{/}n$ and have pairwise distances less than $2\mathrm{/}n$. They are Cauchy and converge by [L2]. Write $\sigma (q_i)={\lim }_j{\sigma }_{m_j}(q_i)$. The only choice was the simultaneous selection of finite covers.

(The limit on the dense set is Lipschitz and extends to $I$.) For $q,q^{\mathrm{\prime }}\in D$, continuity of $d$ and the Lipschitz bound give $d(\sigma (q),\sigma (q^{\mathrm{\prime }}))={\lim }_{j}d({\sigma }_{m_j}(q),{\sigma }_{m_j}(q^{\mathrm{\prime }}))\le K\mathrm{\mid }q-q^{\mathrm{\prime }}\mathrm{\mid }$ [L2]. For $t\in I$ choose a sequence $q_n\in D$ with $q_n\to t$ [L1]; by the previous estimate $(\sigma (q_n){)}_n$ is Cauchy, hence convergent in the complete space $B$, and the limit does not depend on the chosen sequence (two sequences may be interleaved). Define $\sigma (t)$ as this limit; passing to the limit in the Lipschitz estimate shows $\sigma$ is $K$-Lipschitz on $I$.

(Uniform convergence on compact subintervals.) Let $[a,b]\subseteq I$ be a compact subinterval and let $\epsilon >0$. Choose finitely many points $q_1,\dots ,q_N\in D$ such that every $t\in [a,b]$ satisfies $\mathrm{\mid }t-q_i\mathrm{\mid }\le \epsilon \mathrm{/}(3K+3)$ for some $i$; this is possible, including when $a=b$, because $D$ is dense in $I$ and a compact interval is totally bounded in the sense of admitting such finite nets, the case $K=0$ being immediate from the estimator below with $\epsilon \mathrm{/}(3K+3)$ read as any positive number. Since ${\sigma }_{m_j}(q_i)\to \sigma (q_i)$ for each $i$, there is $j_0$ such that $d({\sigma }_{m_j}(q_i),\sigma (q_i))<\epsilon \mathrm{/}3$ for all $j\ge j_0$ and all $i\le N$. For $t\in [a,b]$ pick $i$ with $\mathrm{\mid }t-q_i\mathrm{\mid }\le \epsilon \mathrm{/}(3K+3)$; then $d({\sigma }_{m_j}(t),\sigma (t))\le d({\sigma }_{m_j}(t),{\sigma }_{m_j}(q_i))+d({\sigma }_{m_j}(q_i),\sigma (q_i))+d(\sigma (q_i),\sigma (t))<\frac{\epsilon }{3}+\frac{\epsilon }{3}+\frac{\epsilon }{3}=\epsilon$ for $j\ge j_0$, because both ${\sigma }_{m_j}$ and $\sigma$ are $K$-Lipschitz. Hence ${\sigma }_{m_j}\to \sigma$ uniformly on $[a,b]$.