If $X$ is a locally compact, locally connected, connected metrizable space, does that imply that there must be a metric $d$ on $X$ such that $B(x, r) = \{y\in X : d(x, y) < r\}$ is connected for all $x\in X, r > 0$?
Note that if $X$ admits such metric then it's necessarily locally connected and connected.
If $X$ has at least two points, then we can always find a metric on $X$ with a disconnected open ball by choosing non-empty disjoint open $U, V\subseteq X$, $x\in U\cup V$ and a metric $d$ on $X$ such that $U\cup V$ is open unit ball around $x$. See here.
Edit: Added the assumption of local compactness.
Best Answer
Here is a direct proof that only assumes $X$ is connected, locally connected, and metrizable. $\newcommand{\diam}{\operatorname{diam}}$
Let $X$ be connected and locally connected metrizable, with topology induced by bounded metric $d$ (note we always have such a bounded metric, as we can replace $d$ with $\min(d,1)$). For $x,y\in X$, let $d'(x,y)=\inf_\Gamma \diam(\Gamma)$, where the infimum is taken over connected subsets $\Gamma\subseteq X$ containing $x$ and $y$, and the diameter is taken with respect to $d$.
Then certainly $d'(x,y)\geq d(x,y)$. Moreover, $d'$ is a metric - finiteness comes from connectivity of $X$ and boundedness of $d$, and the triangle inequality follows from the fact that if the connected set $\Gamma_1$ joins $x$ to $y$ and connected $\Gamma_2$ joins $y$ to $z$, then $\Gamma_1\cup \Gamma_2$ joins $x$ to $z$, and has diameter at most $\diam(\Gamma_1)+\diam(\Gamma_2)$.
Finally, if $d(x_n,x)\to 0$, then by local connectivity $d'(x_n,x)\to 0$, so $d$ and $d'$ are equivalent metrics.
But if $y\in B(x,R)$, where the (open) ball is taken with respect to $d'$, then there is a connected set $\Gamma$ joining $x$ to $y$ with $\diam(\Gamma)< R$, and then every $z\in \Gamma$ is also joined to $x$ by a connected set with diameter less than $R$ (namely $\Gamma$), so $d'(z,x)<R$ as well. Hence $\Gamma\subseteq B(x,R)$, and since this applies to all $y\in B(x,R)$, the ball is connected.