No. Consider $X_0 = \{0,1\}$ with the metric inherited from $\mathbb{R}$. Let $X_i = \frac{1}{3}X_{i-1} \cup \left(\frac{1}{3}X_{i-1} + \frac{2}{3}\right)$ again with the metric inherited from $\mathbb{R}$. We are making the the standard Cantor set here with the metric inherited from $\mathbb{R}$. But each $X_i$ has dimension zero being a finite point set, but the standard middle-thirds Cantor set has positive box dimension. Needless to say that all these spaces are compact in the usual topology.

I really should avoid answering questions late at night. My original answer is muddled enough to not work. But here is what it should have been:

Let $X_0 = \{0,1\}$ and $X_i = X_{i-1} \times X_0$. Give each $X_i$ the $2-$adic ultrametric. That is the distance between two sequences of $0'$s and $1'$s is $2^{-n}$ where $n$ is the number of share initial symbols the two sequences share. The projections maps are given by truncation. The inverse limit is now $\mathbb{Z}_2$ which has box counting dimension equal to one.