Upper estimate for diamenter of connected set in $\mathbb R^d$ Let $d \in \mathbb N$ and $|.|: \mathcal P(\mathbb R^d) \to \mathbb [0,\infty]$ be the diameter function, i.e $|A| := \sup \{|x-y|: x,y \in A \}$. Let $B \subset \mathbb R^d$ be any set and $(B_i)_{i \in I}$ be any cover of $B$ for $I$ some index set, i.e we have $B \subseteq \cup_{i \in I} B_i$. At first glance, it might seem reasonable to think that $|B| \leq \sum_{i \in I} |B_i|$ should hold (i.e to think that the diameter function is subadditive). However, one finds easy counterexamples when one either independently allows the index set to be uncountable or $B$ to be not connected. I wonder now if those two assumptions are actually sufficient for the inequality to hold, or in other words, I wonder if the following is true: 
Let $B \subset \mathbb R^d$ be a (path-)connected subset  and $(B_i)_{i=1}^\infty$ a cover of $B$. Then $|B| \leq \sum_{i=1}^\infty |B_i|$.
Edit: I would like to only impose additional conditions on $B$. Hence, if the above claim fails to hold, what else needs to be assumed (at least) of $B$ in order for this to work ? 
 A: The case $d = 0$ is uninteresting, since $\mathbb{R}^0 = \{0\}$ has diameter $0$. So we assume $d \geqslant 1$.
Furthermore, if $L$ is a straight line in $\mathbb{R}^d$, then the orthogonal projection $\pi$ to $L$ is Lipschitz-continuous with Lipschitz constant $1$, hence $\pi(B)$ is a connected subset of $L \cong \mathbb{R}$ (where $\cong$ means an isometry) with $\lvert \pi(B)\rvert \leqslant B$ and $\{\pi(B_i) : i \in \mathbb{N}\setminus \{0\} \}$ is a cover of $\pi(B)$ with $\lvert \pi(B_i)\rvert \leqslant \lvert B_i\rvert$. If $B$ is unbounded, we can find a straight line $L$ such that $\pi(B)$ is unbounded, and if $B$ is bounded, we can find a straight line with $\lvert \pi(B)\rvert = \lvert B\rvert$ (take $x,y\in \overline{B}$ with $\lvert x-y\rvert = \lvert B\rvert$, and let $L$ be the line through $x$ and $y$).
This reduces the problem to the case $d = 1$, and we need only consider $B = [0,t]$. Also, we can replace each $B_i$ with the compact interval $I_i = [\inf B_i, \sup B_i]$ (ignoring all empty $B_i$). And then it's an immediate consequence of the subadditivity and monotonicity of the Lebesgue measure $\lambda$:
$$\lambda(B) \leqslant \lambda\biggl( \bigcup_{i = 1}^{\infty} I_i\biggr) \leqslant \sum_{i = 1}^{\infty} \lambda(I_i) = \sum_{i = 1}^{\infty} \lvert I_i\rvert = \sum_{i = 1}^{\infty} \lvert B_i\rvert.$$
