Loop space of semi-locally simply connected is locally path-connected I came across the following statement (link).

Let $X$ be a locally path-connected and semi-locally simply connected topological space. Denote by $\Omega$ the loop space with basepoint $x\in X$, endowed with the compact-open topology. Then $\Omega$ is locally path-connected.

The page linked above does not provide any proof. Could anyone provide some reference, or give some hints about how to prove this fact (if it helps, for my purposes I can assume $X$ is also simply connected)?
To clarify:


*

*locally path-connected means that $X$ has a basis of path-connected subsets;

*semi-locally simply connected means that $X$ has a basis of subsets $\{U\}$ such that the inclusions $\pi_1(U)\to\pi_1(X)$ are the zero maps (see here for details);

*the loop space with basepoint $x$ is the set of continuous functions $f:[0,1]\to X$ such that $f(0)=f(1)=x$.

 A: It is false.
Let $X$ be the cone on the Hawaiian earring $H = \bigcup_{n=1}^\infty S_n \subset \mathbb R ^2$, where $S_n$ is the circle around $(0,1/n)$ with radius $1/n$. We have $S_n \cap S_m = \{(0,0)\}$ for $n \ne m$. We may write $X = \{ t(0,0,1) + (1-t)(x,y,0) \mid t \in I, (x,y) \in H \} \subset \mathbb R ^3$. The cone point is $(0,0,1)$ and $X$ inherits a metric $d$ from $\mathbb R ^3$. Let $x_0 = (0,0,0) \in X$ the basepoint of $X$.
$X$ is contractible, hence semi-locally simply connected, but it is not locally simply connected locally simply-connected vs. semilocally simply-connected. 
The compact-open topology on $\Omega X$ agrees with the metric topology induced by the $\sup$-metric $d_\infty(\ell,\ell') = \sup \{d(\ell(t),\ell'(t)) \mid t \in I \}$.
Assume that $\Omega X$ is locally path connected.
Consider the constant loop $c(t) \equiv x_0$. Let $W_r = \{ \ell \in X \mid d_\infty(c,\ell) < r \}$. We find an open path connected neighborhhod $W'$ of $c$ such that $W'\subset W_1$ and $r > 0$ such that $W_r \subset W'$. Let $\ell_n$ be the loop in $X$ parametrizing the circle $S_n \times \{ 0 \} \subset X$. Take $n$ such that $1/2n < r$. Then $\ell_n \in W_r$. Choose a path $u : I \to W_1$ such that $u_0 = c, u_1 = \ell_n$. We have $d_\infty(c,u(t)) < 1$, hence the loop $u(t)$ does not go through $\{(0,0,1)\}$. The path $u$ yields a homotopy $u' : c \simeq \ell_n$ such that all $u'_t = u(t)$. By construction $u'$ is a homotopy in $X' = X \setminus \{(0,0,1)\}$. There is a retraction $d : X' \to H$, hence we get $dc \simeq d\ell_n$. But this is not true which shows that $\Omega X$ is not locally path connceted.
Remark:
The link in your question says: In general, if $X$ is locally $n$-connected, $\Omega X$ is locally $(n−1)$-connected. This seems more plausible, although I do not know a proof.
