Space with semi-locally simply connected open subsets 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 exactly one morphism. 

My question is: is it true that if a space $X$ is such that, for any open subset $U\subseteq X$ ($X$ included), $U$ is semi-locally simply connected, then $X$ must be locally simply connected? 

Notice that this implies that $X$ is locally path connected. I tried to post this question on stack exchange some months ago, but I didn't receive any answer. It was just something that came to my mind while I was studying for my master thesis.
 A: $X$ need not be locally simply connected. Consider the following construction:
Let $X_0=S^1$ and $X_n=CS^1$ for all $n\geq 1$ where $CS^1$ is the cone over the circle. Let $Y_0=X_0$, and for $n\geq 1$, let $Y_n\subseteq X_n$ be the base of the cone, $S^1\times\{0\}$ and $Z_n\subseteq X_n$ be the arc $\{b_0\}\times I$ where $b_0$ is the basepoint of $S^1$. Let $X'$ be the shrinking wedge of all $X_n$, that is, every neighborhood of the wedge point contains all $X_n$ with $n\geq k$ for some $k$. Finally, let $X=X'/\sim$, where the relation $\sim$ glues $Z_n$ along $Y_{n-1}$ by the exponential map for all $n\geq 1$.
 
Such a space is not locally simply connected since every neighborhood $U$ of the basepoint must include some $Y_n\approx S^1$ yet not include the cone $X_n$ necessary to nullhomotope a loop around $Y_n$. Yet it is locally semilocally simply connected since we can choose a neighborhood $V\subseteq U$ such that $Y_n\subseteq V$ only if $X_n\subseteq U$. Thus the inclusion $V\to U$ induces a trivial homomorphism on fundamental groups.
