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.
- Using the lifting criterion, why if $Y$ is simply connected, the lifting exists?
This should be understood as "if $Y$ is additionally simply connected (to being locally path connected) then the lifting always exists". And that's because $\pi_1(Y)$ is trivial and thus $f_*(\pi_1(Y))\subseteq p_*(\pi_1(E))$ is always trivially satisfied.
Otherwise it is not necessarily true.
- Does simply connected implies locally pathwise connected? And if it does, do you have a proof?
No, it doesn't. Take $A=\{\frac{1}{n}\ |\ n\in\mathbb{N}\}\cup\{0\}$ and then define $X=(A\times [0,1])\cup([0,1]\times \{0\})$. Also known as the comb space. Note that $X$ is not only simply connected, but even contractible. However it is not locally path connected.
Also, since English is not my native language, I read both the terms locally path connected and locally pathwise connected, do these terms have the same definition or do they refers to different concepts?
These are the same things. And it means that for every point $x\in X$ and every open neighbourhood $U$ of $x$ there is a path connected open neighbourhood $V$ of $x$ such that $V\subseteq U$.
Best Answer
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.