The product $\prod_{i\in\omega}X_{i}$
is the set of all sequences $(x_{0},x_{1},...)$
where $x_{i}\in X_{i}$
for each $i$
.
Now you are given a family of continuous “bonding” maps $f_{i}:X_{i+1}\to X_{i}$
.
The inverse limit is a special subset of $\prod_{i\in\omega}X_{i}$
, consisting of those sequences $(x_{0},x_{1},...)$
satisfying $f_{i}(x_{i+1})=x_{i}$
for each $i$
. I like to think of the elements of the inverse limit as “threads” because the terms are linked together by the functions $f_{i}$
. If you picture this from $X_{0}$
going up, you get a tree-like structure.
Now give this subset the subspace topology, where $\prod_{i\in\omega}X_{i}$ has the product topology. That is, $U$ is open in $X^*$ iff $U=V\cap X^*$ for some open $V$ in $\prod_{i\in\omega}X_{i}$.
First of all, the topology $\prod_i\tau_i$ is not the product topology, but the so-called box topology. They only coincide if the familly is finite. The true product topology is generated by products of open sets $U_i\in\tau_i$ (up to here, this is the box topology), but those open sets must be $U_i=X_i$ except for finitely many $i$'s.
Second, the product topology makes all projections continuous, but this does not characterize it. The property that characterizes it is that a mapping $\ f:Y\to\prod_iX_i$ is continuous if and only if all components $f_i=\pi_i\circ f$ are continuous (note the if and only if).
Next, concerning the questions about inverse images of compact sets, this is essentially related to the notion of proper map. In absolute generality proper means universally closed, that is the mapping is closed and any extension $f\times$Id${}_Y$ is closed too (look at wiki, for instance). However technical this is, under mild restrictions this is equivalent to what you asked for: inverse images of compact sets are compact. Now given $\pi_i:X_i\times Y_i\to X_i$ (denote $Y_i=\prod_{j\ne i}X_j$) and $K\subset X_i$ compact, $\pi_i^{-1}(K)=K\times Y_i$ is compact if and only if (Thychonoff) $Y_i$ is compact, if and only (Tych again) if all $X_j$ are compact. Thus projections are proper if and only if all factors of the product are compact.
Finally suppose the factors $X_i$ compact and also Hausdorff. Consider any continuous mapping $f:X\to X_i$. Then any compact set $K\subset X_i$ is closed (by the Hausdorff assumption), hence $f^{-1}(K)$ is closed in $X$, which is compact (Tych once again). But closed in compact is compact, hence $f^{-1}(K)$ is compact. (This is just the general argument to show that a continuous mapping from a compact space to a Haussdorf one is proper.)
Best Answer
In general it is not closed, but it is closed in the case of the $X_i$ being Hausdorff:
The proof given here is in a way a generalization of the proof of the equivalence of Hausdorff and the closure of the diagonal in the product:
Take $(x_i)_I \in \prod_{i\in I} X \setminus \varprojlim X_i$ then there are $j\preceq k \in I$ such that $\phi_{kj}(x_k)\not=x_j$, as $X_j$ is Hausdorff we can choose disjoint neighborhoods $U_j,V_j\subset X_j$ of $X_j, \phi_{kj}(x_k)$ respectively. Now $V_k = \phi_{kj}^{-1}(V_j) \subset X_k$ is a neighborhood of $x_k$. Therefor $\prod_{i\in I} W_i$ with $W_i = X_i$ for $i\not=k,j$ and $W_k = V_k, W_j = U_j$ is a neighborhood of $(x_i)_I$ and disjoint from $\varprojlim X_i$ as $\phi_{kj}(W_k)$ disjoint from $W_j$. So $\varprojlim X_i$ is closed as the complement of an open set.
As a reference see for example: 1. chapter of Luis Ribes and Pavel Zalesskii. Profinite groups. English. 2nd ed. Vol. 40. Ergeb. Math. Grenzgeb., 3. Folge. Berlin: Springer, 2010. isbn: 978-3-642-01641-7