It is not hard to find an inverse system of connected spaces with the disconnected inverse limit(It's enough to look at $\mathbb R^2$). Now my question is in the following.
Under which conditions the inverse limit of path-connected spaces is path-connected or if somebody knows a paper in that regard, then he is welcome to cite here.
This is only a partial answer. Let us first give an example showing that even under "very nice" conditions an inverse limit of path connected spaces may not be path connected.
Let $S = \{(t, \sin(\frac{1}{t})) \mid t \in (0,1] \} \subset \mathbb{R}^2$ be the topologist'sine curve. Its closure $T$ in $\mathbb{R}^2$ is called the closed topologist'sine curve. We have $T = \{ 0 \} \times [-1,1] \cup S$. Define $$R_n = [-\frac{1}{n}, \frac{1}{n}] \times [-1-\frac{1}{n},1+\frac{1}{n}] \cup S_n ,$$ $$S_n = \{ (t,x) \mid t \in [\frac{1}{n},1], x \in [\sin(\frac{1}{t})-\frac{1}{n},\sin(\frac{1}{t})+\frac{1}{n}]$$ and $T_n = R_n \cup S_n$. It is easy to see that $S_n$ is homeomorphic to $X_n = [\frac{1}{n},1] \times [\sin(n)-\frac{1}{n},\sin(n)+ \frac{1}{n}]$ via a homeomorphism keeping $\{ \frac{1}{n} \} \times [\sin(n)-\frac{1}{n},\sin(n)+ \frac{1}{n}]$ fixed. Hence $T_n$ is homeomorphic to $R_n \cup X_n$ which is obviuosly homeomorphic to a plane disk. We have $T_{n+1} \subset int T_n$ and $\bigcap_{n=1}^\infty T_n = T$. Hence we obtain an inverse system $\mathbf{T} = (T_n,i_n)$ indexed by $\mathbb{N}$ and bonded by inclusions $i_n : T_{n+1} \to T_n$. We have $\varprojlim \mathbf{T} = T$.
Now some positive news. If you make a Google search for "local connectivity of inverse limit spaces" you will find a number of interesting papers like
https://arxiv.org/pdf/1712.03180.pdf
https://www.researchgate.net/publication/321718753_Local_k-connectedness_of_an_inverse_limit_of_polyhedra
Perhaps this is helpful.