A topological space $X$ is semi-locally simply connected if, for any $x\in X$, there exists an open neighbourhood $U$ of $x$ such that any loop in $U$ is homotopically equivalent to a constant one in $X$ or, equivalently, if the functor $$\Pi_1(U)\rightarrow\Pi_1(X)$$ induced by the inclusion $U\subseteq X$ factorizes through a groupoid in which for each pair of objects there is at most one morphism.
My question is: is it true that if a locally path connected space $X$ is such that, for any open subset $U\subseteq X$, $U$ is semi-locally simply connected, then $X$ must be locally simply connected?
Consider the subspace of $\mathbb{R}^2$ that is the union of line segments from $(0,0)$ to $(x,1)$, where $x\in\{0\}\cup\{1/n:n\in\mathbb{Z},n>0\}$. This is simply connected and "locally semilocally simply connected", but it is not locally simply connected because it is not locally path connected. (The definition of simply connected is that $\pi_0$ and $\pi_1$ are trivial.)
Perhaps the hypotheses imply that the space is locally $\pi_1$-trivial -- I could not think of an example that wasn't.
Since locally simply connected implies locally path connected, the above example implies we ought to add locally path connected to the list of hypotheses. This implies every open subspace has a universal cover, but I'm not sure whether this means the space is locally simply connected.