The answer is ``yes'' if $X$ is Hausdorff (by applying a standard technique involving compactness and the local connectedness).

On the other hand, some separation axioms for $X$ are necessary: it may happen that $X$ contains a special point $x^*\in X\setminus C$ contained in any non-empty open set. In this case the sets $V_i,V_j$ cannot be made disjoint.

Such pathological space can be constructing as the set $X=[0,1]\cup\{x^*\}$ where $x^*\notin[0,1]$, endowed with the topology consisting of the empty sets and sets $W\subset X$ that contain $x^*$ and have open intersection $W\cap[0,1]$ in the Euclidean topology of $[0,1]$.

The answer is ``yes'' if $X$ is Hausdorff.

We can identify the arc $C$ with the unit interval $[0,1]$ and assume that $[0,1]$ is contained in $X$ and is covered by a family $\mathcal U$ of open subsets of $X$. By the compactness of $[0,1]$, there exists an increasing sequence $0=a_0<a_1<\dots<a_n=1$ of real numbers such that for every positive $i\le n$ the interval $[a_{i-1},a_{n}]$ is contained in some set $U_i\in\mathcal U$ of the cover. Using the local connectedness and the Hausdorff property of $X$, chose a sequence $(W_i)_{i=0}^n$ of pairwise disjoint open connected sets in $X$ such that $a_i\in W_i\subset U_i\cap U_{i-1}$ for all $i\in\{0,\dots,n\}$. Here we assume that $U_{0}=X$.

For every $0<i\le n$ find a compact connected subset $K_i$ in the open interval $(a_{i-1},a_{i})$ such that $[a_{i-1},a_i]\subset W_{i-1}\cup K_i\cup W_i$ and observe that the sets $K_i$, $0<i\le n$, are pairwise disjoint. Using the Hausdorff property of $X$ and the compactness of the pairwise disjoint sets $K_i$, we can find a sequence $(O_i)_{i=1}^n$ of pairwise disjoint open sets of $X$ such that $K_i\subset O_i\subset U_i$ for every $i\le n$. Moreover, since $X$ is locally connected, we can additionally assume that each open neighborhood $O_i$ of the connected set $K_i$ is connected. Finally for every positive $i\le n$ consider the open connected neighborhood $V_i=W_{i-1}\cup O_i\cup W_i$ of the segment $[a_{i-1},a_i]$ and observe that $V_i\subset U_i\in\mathcal U$. Moreover, for any $0\le i<j\le n$ if $j-i>1$, then the sets $V_i,V_j$ are disjoint and if $j-i=1$, then $V_i\cap V_j=W_i$ is conneceted.

On the other hand, some separation axioms for $X$ are necessary: it may happen that $X$ contains a special point $x^*\in X\setminus C$ contained in any non-empty open set. In this case the sets $V_i,V_j$ cannot be made disjoint.

Such pathological space can be constructing as the set $X=[0,1]\cup\{x^*\}$ where $x^*\notin[0,1]$, endowed with the topology consisting of the empty sets and sets $W\subset X$ that contain $x^*$ and have open intersection $W\cap[0,1]$ in the Euclidean topology of $[0,1]$.

The answer is ``yes'' if $X$ is Hausdorff (by applying a standard technique involving compactness and the local connectedness).

On the other hand, some separation axioms for $X$ are necessary: it may happen that $X$ contains a special point $x^*\in X\setminus C$ contained in any non-empty open set. In this case the sets $V_i,V_j$ cannot be made disjoint.

Such pathological space can be constructing as the set $X=[0,1]\cup\{x^*\}$ where $x^*\notin[0,1]$, endowed with the topology consisting of the empty sets and sets $W\subset X$ that contain $x^*$ and have openintersection $W\cap[0,1]$ in the Euclidean topology of $[0,1]$.

The answer is ``yes'' if $X$ is Hausdorff (by applying a standard technique involving compactness and the local connectedness).

On the other hand, some separation axioms for $X$ are necessary: it may happen that $X$ contains a special point $x^*\in X\setminus C$ contained in any non-empty open set. In this case the sets $V_i,V_j$ cannot be made disjoint.

Such pathological space can be constructing as the set $X=[0,1]\cup\{x^*\}$ where $x^*\notin[0,1]$, endowed with the topology consisting of the empty sets and sets $W\subset X$ that contain $x^*$ and have open intersection $W\cap[0,1]$ in the Euclidean topology of $[0,1]$.