Extension of a Uniformly Continuous Real Function from a Dense Subset

In the setting of The Real Numbers: Standing Notation and Background, let $(X,d)$ be a metric space, whose open sets form a topology $\mathcal{T}_{d}$ on $X$ by The Open Subsets of a Metric Space Form a Topology, and let $S\subseteq X$ be nonempty and dense in $(X,\mathcal{T}_{d})$. For a real-valued function defined on a subset of $X$, being continuous on that subset and being uniformly continuous on that subset are as defined there, taken with respect to the metric $d$ and the metric $d_{\mathbb{R}}$ of The Real Numbers: Standing Notation and Background (i) on $\mathbb{R}$. Then the following hold.

1. (Existence of a uniformly continuous extension)(i) Let $f:S\to\mathbb{R}$ be uniformly continuous on $S$. Then there is a function $g:X\to\mathbb{R}$ that is uniformly continuous on $X$ and continuous on $X$, and that satisfies $g(x)=f(x)$ for every $x\in S$.

2. (A continuous function is determined on a dense subset)(ii) If $g_{1},g_{2}:X\to\mathbb{R}$ are continuous on $X$ and satisfy $g_{1}(x)=g_{2}(x)$ for every $x\in S$, then $g_{1}(x)=g_{2}(x)$ for every $x\in X$.

3. (Moduli pass from a dense subset)(iii) Let $g:X\to\mathbb{R}$ be continuous on $X$ and let $\varepsilon,\theta\in\mathbb{R}$ be positive. If all $x,y\in S$ with $d(x,y)\le\theta$ satisfy $|g(x)-g(y)|\le\varepsilon$, then all $x,y\in X$ with $d(x,y)<\theta$ satisfy $|g(x)-g(y)|\le\varepsilon$.

4. (Bounds pass from a dense subset)(iv) Let $g:X\to\mathbb{R}$ be continuous on $X$ and let $C\in\mathbb{R}$ satisfy $|g(x)|\le C$ for every $x\in S$. Then $|g(x)|\le C$ for every $x\in X$.