Piecewise linear functions are not dense in $W^{1,\infty}(\Omega;\mathbb R^n)$ for any open set $\Omega\subset\mathbb R^m$.
If it were true for $\Omega$, it would also be true for any open subset, in particular any open ball. Recall that the Sobolev space $W^{1,\infty}(\Omega;\mathbb R^n)$ is the space of Lipschitz functions $\Omega\to\mathbb R^n$ with the same norm if $\Omega$ is quasiconvex — and balls are quasiconvex. (This and many other basic facts of Lipschitz functions can be found in Lectures on Lipschitz analysis by Heinonen.) The Lipschitz condition behaves well when restricting to subspaces, which is not that obvious from the point of view of Sobolev spaces.
If piecewise linear functions were dense, for any $f\in W^{1,\infty}$ there would be a sequence of piecewise linear functions $f_i$ tending to $f$. Every Lipschitz function on a line can be extended to a Lipschitz function on the whole domain and the restriction of a piecewise linear function to a line is piecewise linear. Therefore, if we restrict all functions to a line segment in $\Omega$ and evaluate only one component, we end up with the claim that every Lipschitz function $[a,b]\to\mathbb R$ can be approximated by a piecewise linear function in the Lipschitz norm.
Now, let $f:[a,b]\to\mathbb R$ be an arbitrary Lipschitz function and suppose there is a sequence of piecewise linear functions $f_i:[a,b]\to\mathbb R$ converging to it. Then $f_i(a)\to f(a)$ and $f_i'\to f'$ in $L^\infty$. Since $f'$ can be any $L^\infty$ function, this means that piecewise constant functions are dense in $L^\infty$. But this is false. The function $g:[0,1]\to\mathbb R$, $g(x)=\sum_{k=1}^\infty\chi_{[2^{-2k},2^{1-2k}]}(x)$ is in $L^\infty$ but any piecewise constant function has at least distance $\frac12$ from it.