Let $(X, \mathcal{T})$ be the pseudoarc, which is a hereditarily indecomposable continuum. Here hereditary means on every subcontinuum. Subcontinuums of a continuum are exactly its closed and connected subsets.
In a previous question it was established that any indecomposable continuum is nowhere locally connected. Hence the pseudoarc is nowhere locally connected on any closed and connected subspace.
Suppose $U \subset X$ is non-empty and $(U, \mathcal{T}|U)$ does not contain isolated points.
Is $(U, \mathcal{T}|U)$ nowhere locally connected?
This is a partial answer which provides an affirmitive answer for the non-empty locally closed subsets without isolated points.
If $U \in \mathcal{T} \setminus \{\emptyset\}$, then $\mathcal{T}|U$ is nowhere locally connected. For otherwise $\mathcal{T}$ would be not be nowhere locally connected.
The following shows a that each non-empty closed subset which does not contain a singleton component is nowhere locally connected. For let $V \subset X$ be such. Then the connected components of $V$ are closed, connected, and non-degenerate, hence pseudoarcs. Now let $x \in V$ and $U \in (\mathcal{T}|V)(x)$ be connected. By connectedness, $U$ must be contained in some component of $C$. Hence $V$ is not locally connected at $x$.
The following extends the previous to show that any non-empty closed subset without isolated points is nowhere locally connected. For if some component of $V$ is a singleton $\{x\}$, then there is no connected open subset which contains $x$, since that would imply that $\{x\}$ is open and hence isolated.
A corollary is that any non-empty locally closed subset without isolated points is nowhere locally connected. Here locally closed means an intersection of an open set and a closed set.