Note that if we have a sequence of $u_{n} \in C^{1}(\Omega)$ converging to some function $u$ with respect to the $||\;||_{W^{1,\infty}}(\Omega)$ - norm, then by completeness of the space $X = (C^{1}(\Omega),||\;||_{W^{1,\infty}(\Omega)})$, the limit $u$ is also in $X$.
But $W^{1,\infty}(\Omega)$ does not only contain $C^{1}$-functions - it is easy to construct a Lipschitz function that does not have a continuous derivative.
Thus, the statement is wrong - we can´t even hope to approximate by $C^{1}$-functions.
Note that the other questions you linked handle a slightly different (weaker) problem:
$||\nabla{u_{n}}|| \rightarrow ||\nabla{u}||$ does not imply that $$||\nabla{u}-\nabla{u_{n}}|| \rightarrow 0$$
The answer is negative for general open sets; to see this let $\Omega = B_1 \setminus \{0\} \subset \Bbb R^n$ be the punctured ball in $\Bbb R^n,$ and take $p < n < q.$ Fix any $\varphi \in C^{\infty}_c(B_1)$ such that $\varphi(0)=1.$
By Sobolev embedding convergence in $W^{1,q}$ implies uniform convergence, so $\varphi \in W^{1,q}(\Omega) \setminus W^{1,q}_0(\Omega).$ On the other hand I claim $\varphi \in W^{1,p}_0(\Omega)$ by considering
$$ v_r(x) = 1 - \frac{\int_{|x|}^1 t^{(1-n)/(p-1)} \,\mathrm{d} t}{\int_r^1 t^{(1-n)/(p-1)} \,\mathrm{d} t} $$
for $0 < r < |x| < 1,$ setting $v_r(x) = 0$ for $|x| \leq 1.$ Then $v_r \in C^{1}_c(\Omega)$ and one can compute that
$$ \int_{B_1} |\nabla v_r(x)|^p \,\mathrm{d} x = \omega_{n-1} \left(\frac{p-1}{n-p}\right)^{1-p} |1 - r^{(p-n)/(p-1)}|^{1-p} \to 0$$
as $r \to 0.$ Hence $v_r \to 1_{\Omega}$ in $W^{1,p}(\Omega),$ and by mollifying $v_r \varphi$ we see that $\varphi \in W^{1,p}_0(\Omega)$ as required.
The above counterexample is furnished using the theory of Sobolev capacity, and I've adapted the calculations from 2.11 of the following reference.
Heinonen, Juha; Kilpeläinen, Tero; Martio, Olli, Nonlinear potential theory of degenerate elliptic equations, Oxford Mathematical Monographs. Oxford: Clarendon Press. v, 363 p. (1993). ZBL0780.31001.
More generally from chapters 2 and 4 of the text we have the following results, which provide some insight about when we may or may not expect results of this type to hold true.
Theorem (2.43): If $\Omega \subset \Bbb R^n$ is open and $E \subset \Omega$ is relatively closed, then we have
$$ W_0^{1,p}(\Omega) = W^{1,p}_0(\Omega \setminus E) $$
if and only if $E$ has $p$-capacity zero.
Theorem (4.5): If $\Omega \subset \Bbb R^n$ is open, then $\varphi \in W^{1,p}(\Omega)$ lies in $W^{1,p}_0(\Omega)$ if and only if there is a $p$-quasicontinuous function $\tilde \varphi$ on $\Bbb R^n$ agreeing with $\varphi$ almost everywhere in $\Omega,$ such that $\tilde\varphi = 0$ $p$-quasieverywhere in $\Bbb R^n \setminus \Omega.$
Using Theorem 2.43 we can generate more counterexamples as follows: suppose $E \Subset \Omega$ has $p$-capacity zero but non-zero $q$-capacity, and choose $\varphi \in C_c^{\infty}(\Omega)$ such that $\varphi \equiv 1$ on $E.$ Then we see that $\varphi \not\in W^{1,q}_0(\Omega \setminus E),$ but $\varphi \in W^{1,p}_0(\Omega \setminus E) = W^{1,p}_0(\Omega).$ The above counterexample does this by noting the point $\{0\}$ has $p$-capacity zero if and only if $p < n,$ reproducing the necessary calculations.
Best Answer
By the fundamental theorem of calculus we have that, if $u\in C_c^\infty(\mathbb{R}^n)$ and $x_0$ is such that $u(x_0)=0$ then
$$ u(x)=\int_0^1 \nabla u(tx+(1-t)x_0)\cdot (x-x_0)dt $$
from which we get
$$ |u(x)|\leq \| \nabla u \|_\infty |x-x_0| $$
Now if $\Omega$ has finite width and $u\in C_c^\infty(\Omega)$, then for every $x\in \Omega$ there is an $x_0 \notin \Omega$ such that $|x-x_0| \leq D$ (where $D$ is the distance between the two parallel hyperplanes bounding $\Omega$). We conclude that
$$ \| u\|_{\infty} \leq D \| \nabla u\|_{\infty}. $$