For any topological space $X$, the suspension $\Sigma X$ is path-connected and hence connected. To see this, suppose $[(x_1, a_1)], [(x_2, a_2)] \in \Sigma X$ where the square brackets denote the equivalence class of a pair from $X \times [0, 1]$ under the equivalence relation $\sim$.
The path $p_1 : [0, 1] \to \Sigma X$, $p_1(t) = [(x_1, (1-t)a_1 + t)]$ joins $p_1(0) = [(x_1, a_1)]$ and $p_1(1) = [(x_1, 1)]$.
The path $p_2 : [0, 1] \to \Sigma X$, $p_2(t) = [(x_2, 1 - t + ta_2)]$ joins $p_2(0) = [(x_2, 1)]$ and $p_2(1) = [(x_2, a_2)]$.
As $p_1(1) = [(x_1, 1)] = [(x_2, 1)] = p_2(0)$, the concatenation of the paths $p_1$ and $p_2$ gives a path $p := p_1\ast p_2$ joining $[(x_1, a_1)]$ and $[(x_2, a_2)]$.
On the other hand, $\Sigma X$ need not be simply connected. For example, if $X = S^0 = \{-1, 1\}$ (equipped with the discrete topology), then $\Sigma S^0 = S^1$ which is not simply connected. In general, $\Sigma S^n = S^{n+1}$. However, if $X$ is path-connected, then $\Sigma X$ is simply connected; this follows from the Seifert-van Kampen Theorem.
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
It is not true. Take any connected but not pathwise connected well-pointed $(X,*)$. Since $(X,*)$ is well-pointed, the reduced suspension has the same homotopy type as the unreduced suspension $SX$. We shall show that $SX$ is not simply connected which implies that $\Sigma X$ is not simply connected.
Working with reduced singular homology, we know that $H_1(SX) = \tilde{H}_1(SX) \approx \tilde{H}_0(X)$. See Reduced Homology on unreduced suspension . Since $X$ is not pathwise connected, $H_0(X)$ is a free abelian group with more than one generator, hence $\tilde{H}_0(X) \ne 0$ and thus $H_1(SX) \ne 0$.
The first homology group is the the abelianization of the fundamental group (see The First Homology Group is the Abelianization of the Fundamental Group.}). Hence the fundamental group of $SX$ cannot be trivial.