Here it is stated, that for a Lie-Algebra Bundle the isomorphism-class of the Lie-Algebra is locally constant, if the Lie-Algebra is rigid. Especially it is stated, that semisimple Lie-Algebras are rigid.
Consider now the following family of Lie-Brackets on $\mathbb{R}^3$ defined via its action on the basis elements $X_1, X_2, X_3$:
\begin{align} [X_1, X_2]_\epsilon &= \epsilon^2 X_3 \\ [X_2, X_3]_\epsilon &= X_1 \\ [X_3, X_1]_\epsilon &= X_2 \end{align}
For all $\epsilon > 0$ this Lie-Algebra is isomorphic to $\mathfrak{so}(3)$, which is rigid, since it is semisimple. But as $\epsilon \rightarrow 0$ one gets the Lie-Algebra $\mathfrak{iso}(2)$ of the 2-dimensional Euclidean group. Further one can interpret this limit as a Wigner-Inönü Contraction of $\mathfrak{so}(3)$ (see here for both claims) and as far as I understand, $\mathfrak{so}(3)$ is not isomorphic to $\mathfrak{iso}(2)$.
Now one could consider the smooth, trivial vector bundle $P = \mathbb{R} \times \mathbb{R}^3$ and endow it with a Lie-algebra bundle structure
\begin{equation} \theta: P \times_\mathbb{R} P \rightarrow P \end{equation}
via
\begin{equation} \forall \epsilon \in \mathbb{R}: \forall X, Y \in P_\epsilon: \theta(X, Y) = [X, Y]_\epsilon \end{equation}
where $P_\epsilon = \{\epsilon\} \times \mathbb{R}^4$ denoes the fiber of $P$ over $\epsilon \in \mathbb{R}$.
Here the Lie-Algebra over $0$ is given by $\mathfrak{iso}(3)$ but elsewhere it is given by $\mathfrak{so}(3)$. Hence the isomorphism class is not locally constant. But $\mathfrak{so}(3)$ is rigid, and hence [1] says, that this situation should not occur.
Where I am missing something? I am not trained in this branch of mathematics. I ask the question here, since I want to grasp the point faster.