No. There are only very few general results about uncountable inductive limits. The following example is stated (without proof) in an article of Komura [Some examples on linear topological spaces. Math. Ann. 153 (1964), 150–162]: 

For an uncountable set $I$ and every countable $J\subseteq I$ let $E_J=\{f:I\to \mathbb R: $supp$(f)\subseteq J\}$ (where the support is just $\{i\in I: f(i)\neq 0\}$) endowed with the Frechet topology of pointwise convergence ($E_J$ is isomorphic to $\mathbb R^J$ with the product topology). Then $E=\lim\limits_\to E_J$ is a strict inductive limit and the limit topology is the relative topology of $\mathbb R^I$ (every neighbourhood of $0$ in $E$ contains the absolutely convex hull of $\bigcup_J U_J$ with $0$-neighbourhoods $U_J$ of $0$ in $E_J$ which only give conditions on values $f(j)$ for $j\in F(J)$ *finite*, then there is a finite subset $F$ of $I$ with $I\setminus F \subseteq \bigcup_J J\setminus F(J)$).

Now $K=\{\delta_x: x\in I\}$ is precompact in $E$ (because it is contained in $E$ and precompact in $\mathbb R^I$) but not contained in any step $E_J$.