In section 4.8 of Calegari's book "Foliations and the Geometry of 3-Manifolds", one finds the following quote (bolded emphasis my own):
As we have shown, for $\mathcal F$ taut, leaves of $\tilde{\mathcal F}$ are properly embedded in $\tilde M$; that is, for each leaf $\lambda$ of $\tilde F$, the intersection of $\lambda$ with any compact $K \subset \tilde M$ is compact. We may reinterpret this properness in terms of a comparison between the intrinsic metric $d_\lambda$ in $\lambda$, and the extrinsic metric $d_{\tilde M}|_{\lambda}$ in $\tilde M$. Namely, the properness of the embedding of $\lambda$ in $\tilde M$ is equivalent to the existence of a proper function $$f \colon \lambda \times \mathbb R^+ \to \mathbb R^+$$ which induces an increasing homeomorphism $f(p, \cdot) \colon \mathbb R^+ \to \mathbb R^+$ for each $p \in \lambda$, such that for any two points $p,q \in \lambda$, we have an estimate $$d_\lambda(p,q) \le f(p,d_{\tilde M}(p,q))$$ Informally, for any radius $r$, the ball $B_r^{\tilde M}(p) \subset \tilde M$ intersects $\lambda$ in a set which is contained in the ball $B_{f(r)}^\lambda(p) \subset \lambda$. Here for each $X$, $B_r^X(p)$ denotes the ball of radius $r$ about $p$ in $X$, with respect to the geodesic path metric on $X$.
I am struggling to see how to prove the line in bold, namely the equivalence of $\lambda \hookrightarrow \tilde M$ being a proper embedding with the existence of a proper function as described (in particular, I do not see how to get an increasing homeomorphism), nor do I understand its connection with the informal description near the end.
Is this perhaps a special case of some general result concerning maps between metric spaces restricted here to the inclusion map, e.g. something akin to "if $\phi \colon (X,d_X) \to (Y,d_Y)$, then $\phi$ is a proper map iff there exists $f \colon X \times \mathbb R^+ \to \mathbb R^+$ which induces an increasing homeomorphism $f(p, \cdot) \colon \mathbb R^+ \to \mathbb R^+$ for each $p \in X$, such that for any two points $p,q \in X$, we have $d_X(p,q) \le f(p,d_Y(\phi(p),\phi(q)))$"?
EDIT:
A colleague and I proposed the following proof. Would someone be willing to check it?
Suppose $X, Y$ are proper. Define $B = B(\phi(p),t)$ by $B := \{y \in Y \mid d_Y(y,\phi(p)) \le t \}$. This is closed and bounded, so is compact (by the HB property). Since $\phi$ is proper, the preimage $\phi^{-1}(B)$ is also compact. Define $f \colon X \times \mathbb R^+ \to \mathbb R^+$ by $f(p,t) := \max_{x \in \phi^{-1}(B)} d_X(p,x)$. (The maximum is defined and finite by compactness). Then: $\phi(q) \in B_{d_Y(\phi(p), \phi(q))}(\phi(p))$ so that $q \in \phi^{-1}(B_{d_Y(\phi(p), \phi(q))}(\phi(p)))$ so $$\begin{align*} d_X(p,q) &\le \max_{x \in \phi^{-1}(B_{d_Y(\phi(p), \phi(q))}(\phi(p)))} d_X(p,x) = f(p,d_Y(\phi(p),\phi(q))) \end{align*}$$ Clearly $f_p := f(p, \cdot)$ is increasing with $f(0) = 0$, and an inverse can be defined by flipping the roles of $X$ and $Y$ in the definitions of $B$ and $f_p$, so $f_p \colon \mathbb R^+ \to \mathbb R^+$ is an increasing homeomorphism.
For the converse, suppose a proper function $f$ for $\phi$ exists. We know that if $X,Y$ are proper and $\phi \colon X \to Y$ is continuous, then $\phi$ is proper iff $g(p_n)$ bounded implies $x_n$ bounded for every sequence $\{p_n\}$ in $X$. With this in mind, consider a sequence $\{p_n\} \subseteq X$ such that $\phi(p_n)$ is bounded. Let $q \in X$. Then boundedness of $\phi(p_n)$ implies boundedness of $d_Y(\phi(q),\phi(p_n)$, which implies boundedness of $f(p_n,d_Y(\phi(q),\phi(p_n))$. Since $f(p_n,d_Y(\phi(q),\phi(p_n)) \le f(p_n,d_Y(\phi(q),\phi(p_n))$, this implies $\sup_{n \in \mathbb N} d(p_n,q) < \infty$, so $\{p_n\}$ is bounded.
