Suppose we have a (co)filtered digaram $\dots \rightarrow X_{2}\rightarrow X_{1}$ of topological space. Is is true that the natural map $\pi_{0}[\lim X_{i}]\rightarrow \lim \pi_{0}(X_{i})$ is an isomorphism ?

7No. For example, take all your $X_i$ to be the $1$sphere, and all the maps the degree $2$ map. The limit is actually not pathconnected. – Achim Krause Dec 6 at 21:20
This is not true, for two distinct reasons.
The first is that the inverse system of spaces may not behave well homotopytheoretically. If $X_n = [n, \infty) \subset \Bbb R$, then the limit of $\dots \to X_2 \to X_1 \to X_0$ is empty. However, on path components it is the constant system $\dots \to * \to * \to *$, with limit $*$. Roughly, you might have a path component that is represented by any space $X_i$ that is not represented by any compatible system of points. This might make $\pi_0 \lim X_i \to \lim \pi_0 X_i$ not surjective.
The second is the opposite: the map $\pi_0 \lim X_i \to \lim \pi_0 X_i$ may not be injective. In this case, you may have two points $x$ and $y$ in the limit such that the images in any individual $x_i$ are connected by a path, but where no path can be compatibly lifted all the way up the tower. For example, if $f:S^1 \to S^1$ is a degree2 covering map, then the limit of the tower $$\dots \xrightarrow{f} S^1 \xrightarrow{f} S^1 \xrightarrow{f} S^1$$ is called the 2adic solenoid and it has uncountably many path components.
The first problem goes away if the maps $X_i \to X_{i1}$ are fibrations, and in this case we often call the limit a homotopy limit. The second problem does not go away in this case, but Milnor proved that $\pi_0 \lim X_i$ is built out of two terms: $\lim(\pi_0 X_i)$ and a second term called $\lim^1(\pi_1 X_i)$. In particular, if the spaces $X_i$ are simplyconnected then there are no contributions from the second term, and so there will be an isomorphism $\pi_0(\lim X_i) \to \lim \pi_0(X_i)$.