No smooth non-integrable distribution can be $C^0$ approximated by integrable ones.
For example, consider the following 2-dimensional distribution in $mathbb R^3$: the plane at $(x,y,z)inmathbb R^3$ is spanned by vectors $(1,0,0)$ and $(0,1,x)$. Perturb this distribution within a small $C^0$ distance $varepsilonll 1$. Consider the square in $mathbb R^2$ with vertices $(0,0)$, $(1,0)$, $(1,1)$, $(0,1)$ and let $gamma$ be its boundary (counter-clockwise). This $gamma$ has a "lift", that is a curve $tildegamma$ in $mathbb R^3$ which is tangent to the distribution and whose projection to the horizontal plane is $gamma$. The lift is found by solving an o.d.e., so it is unique if the distribution is smooth but may be non-unique if it is only $C^0$. In the non-perturbed case, the unique lift ends at $(0,0,1)$, hence in the perturbed case all lifts end near $(0,0,1)$. This implies that the distribution is not integrable - if it was integrable, there would be at least one lift (the one contained in a leaf of a foliation) that ends near the origin.
The proof in the general case is similar.
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