Lemma: Given the commutative diagram
$$\begin{array}{ccccccccc} \widetilde{X} & \\
\downarrow{\small{p}} & {\searrow}^{q} \\
X_1 & \!\!\!\!\! \xleftarrow{p_1} & \!\!\!\! X_2\end{array}$$
where $p_1, p$ are covering maps, then so is $q$, where $X_1, X_2, \tilde{X}$ are all path-connected and locally path-connected.
Proof:
$q$ is surjective: $\sigma$ be a path in $X_2$ from $x_0$ and $x$. Pushforward by $p_1$ to get a path $p_1 \circ \sigma$ in $X_1$ from $p_1(x_0) = x_0'$ to $p_1(x)$. Lift to $\widetilde{X}$ to get a path $\widetilde{\sigma}$ starting from some point $x_0''$ in the fiber over $x_0'$. Pushforward by $q$ to get path $q \circ \widetilde{\sigma}$ starting at $x_0$. Uniqueness of path-lifing says $q \circ \widetilde{\sigma} \simeq \sigma$, so that $q$ maps the endpoint of $\tilde{\sigma}$ to the endpoint $x$ of $\sigma$. As $X_2$ is path connected, we can apply this argument for all $x \in X_2$ to prove $q$ is surjective.
$q$ is a covering map: Pick $x \in X_2$. Pushforward by $p_1$ to get $p_1(x)$ in $X_1$. There is a path-connected neighborhood $\mathscr{U}$ of $p_1(x)$ evenly covered by $p_1$ and $p$ (take neighborhoods evenly covered by $p_1$ and $p$ and take intersection). $\mathscr{V}$ be the slice in $p_1^{-1}(\mathscr{U})$ containing $x$. $\{\mathscr{U}_\alpha\}$ be the slices in $p^{-1}(\mathscr{U})$. $q$ maps each slice $\mathscr{U}_\alpha$ to distinct slices in ${p_{1}}^{-1}(\mathscr{U})$. $q^{-1}(\mathscr{V})$ is then union of slices in $\{\mathscr{U}_\alpha\}$ which are mapped homeomorphically onto $\mathscr{V}$. I claim all $\mathscr{U}_\alpha$ are mapped homeomorphically on $\mathscr{V}$ by $q$. This can be proved slicewise, recalling that given a commutative diagram with any two arrows as homeomorphism, so is the third. $\blacksquare$
If $\widetilde{X}$ is simply connected, $p : \widetilde{X}\to X$ the universal cover, $p_1 : X_2 \to X_1$ a covering map, then as $p_*(\pi_1(\widetilde{X}))$ fits inside ${p_1}_*(\pi_1(X_2))$, being the trivial group, we can lift $p$ to $\tilde{p} : \widetilde{X} \to X_1$. By previous discussion, $\tilde{p}$ is a covering map, since it fits inside a commutative diagram like above. Thus, $\widetilde{X}$ covers $X_2$, as desired.
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.
Best Answer
Assuming $\tilde X$ is connected. So your condition shows that $ \pi_1(\tilde X,\tilde x)=0 \ \forall \ \tilde x\in \tilde X$.
I shall show if $x_0\in X$ and $\tilde x_0$ lies in the fiber over $x_0$ of $\tilde X\xrightarrow{p}X$, then $ \pi_1(\tilde X,\tilde x_0)=0\implies X$ is semi-locally simply connected at $x_0$.
Choose an open path-connected neighbourhood $U\ni x_0$ and let $\tilde U\ni \tilde x_0$ such that $p:\tilde U\rightarrow U$ is an isomorphism. Then we have the commutative diagram
$$\require{AMScd} \begin{CD} \tilde U @>\tilde i>> \tilde X\\ @Vp|_{\tilde U}VV @VpVV \\ U @>i>> X \end{CD} $$ Applying $\pi_1$ we get the following commutative diagram $$\require{AMScd} \begin{CD} \pi_1(\tilde U,\tilde x_0) @>\tilde i_*>> \pi_1(\tilde X,\tilde x_0)\\ @V(p|_{\tilde U})_*VV @Vp_*VV \\ \pi_1(U,x_0) @>i_*>> \pi_1(X,x_0) \end{CD} $$
Thus we get from the commutativity $i_*(p|_{\tilde U})_*=p_*\tilde i_*=0$ since $\pi_1(\tilde X,\tilde x_0)=0$
Since $(p|_{\tilde U})_*$ is an isomorphism, we get $i_*=0$
Applying this argument to other points completes the proof.
Edit: For the general case, let $\tilde X=\bigsqcup_i \tilde X_i$ be the connected components of $\tilde X$. Say $\tilde x_0\in \tilde X_{i_0}$ Then $p|_{\tilde X_{i_0}}:\tilde X_{i_0}\rightarrow X$ is a covering map with $\tilde X_{i_0} $ connected and you are back to the previous case.