Let $(S, \rho)$ be an arbitrary metric space. Define $f: S \rightarrow \mathbb{R}^+$.
I have a mathematical object in the form of $Lip(f) diam(S)$ where $$\Lambda(f) = Lip(f)=\sup_{x,y \in S} \frac{|f(x)-f(y)|}{\rho(x,y)}$$ $$D(S) = diam(S) = \sup_{x,y \in S} \rho(x,y)$$
We can rewrite $\Lambda(f) D(S)$ as $\Lambda(f) *\mathcal{H}^1_{D(S)}(S)$ where $$\mathcal{H}^d_{\delta}=inf[\sum_{i=1}^\infty\ D(U_i)^d : \bigcup_{i=1}^{\infty} U_i \supseteq S, D(U_i) \leq \delta ]$$
How can we upperbound this object? Is there an inequality in the form $$\mathcal{H}^1_{\delta}(S) \leq c(\delta) \lim_{\epsilon \rightarrow 0}\mathcal{H}^1_\epsilon(S)$$
It's not true that $D(S)=\mathcal H_{D(S)}^1(S)$. For example if $S=[0,1]\cup[2,3]$ then $D(S)=3$ while $\mathcal H_3^1(S)=2$. (I think it's clearly true if $S$ is connected.)
The inequality you ask about is trivial, with $c(\delta)=1$. If $A\subset B$ then $\inf A\ge\inf B$; hence if $0\le\epsilon\le\delta$ then $\mathcal H_\delta^1(S)\le\mathcal H_\epsilon^1(S)$.