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 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 $E_J$.

EDIT. A related example (certainly folklore, but I don't know about a reference): Again $I$ is an uncountable set and for each countable $J\subseteq I$ we set $E_J=\ell^1(J)$ considered as a subspace of $\ell^1(I)=\{(x_i)_{i\in I}\in\mathbb R^I: \sum_{i\in I}|x_i|<\infty\}$ (where components outside $J$ are $0$). Then $E_J$ is a strict inductive spectrum of Banach spaces and $E=\lim\limits_\to E_J=\ell^1(I)$ *topologically* (the identity $E\to \ell^1(I)$ is continuous because of the universal property of inductive limits, and the identity $\ell^1(I)\to E$ is sequentially continuous because every sequence in $\ell^1(I)$ is contained in a single $E_J$, since $\ell^1(I)$ is metrizable, sequential continuity implies continuity). The unit ball of $\ell^1(I)$ is bounded but not contained iny step. The precompact sets however are indeed contained in a step.

EDIT II

Let me try to make the coincidence of the product topology from $\mathbb R^I$ and the inductive limit topology a bit clearer: We have to estimate every continuous semonorm $p$ on $E$ by $cq_F$ with a constant $c$ and $q_F(f)=\max\{|f(j)|: j\in F\}$ for some finite set $F$ (these are the standard seminorms on the product). By definition of the inductive limit the restriction of $p$ to each $E_J$ can be estimated by $c_J q_{F(J)}$ with some finite set $F(J)\subseteq J$, in particular, $p(f)=0$ for every $f\in E_J$ with supp$(f)\cap F(J)=\emptyset$.

I claim that for the family $(F(J))_{J \text{ countable}}$ there is a finite set $F$ such that $I\setminus F \subseteq \bigcup_J J\setminus F(J)$. Otherwise we find a countable and infinite subset $K$ of $I$ which is not in that union which contradicts the fact that $K\setminus F(K)$ is trivially contained there.

Next, let us see that we have $p(f)=0$ for every $f\in E$ with supp$(f)\cap F=\emptyset$. Let $K$ be the support of $f$. For each $k\in F(K)$ there is $J_k$ such that $k\in J_k\setminus F(J_k)$. Using this we can write $f=g+\sum_{k\in F(K)} f_k$ such that $g\in E_K$ vanishes on $F(K)$ and $f_k\in E_{J_k}$ vanishes on $F(J_k)$ which together with the subadditivity of $p$ yields $p(f)=0$.

Finally, since $E_F$ is finite dimensional we find a constant $c$ such that $p(f)\le c q_F(f)$ for $f\in E_F$ which yields (again by decomposing $f\in E$ as $f=g+h$ with $g\in E_F$ and $h$ vanishing on $F$) that $p(f)\le c q_F(f)$ for all $f\in E$.