I wish to show that the function $x\mapsto 1$ does not lie in the Sobolev space $W_0^{1,1}((-1,1))$ directly, without invoking the Trace Theorem.
I have the "picture" by which the approximating smooth functions would have to lie very close to $1$ along their support, and then decay to zero very quickly to remain supported inside the interval $(-1,1)$. Thus, the closer the functions to 1 in $L^1$ norm, the faster they have to drop off at the boundary, hence the larger the $L^1$ norm of the derivative grows.
But I am having trouble formalising this idea for an arbitrary sequence of smooth, compactly supported functions converging to $1$ in $W^{1,1}$ norm.
Hint: for a smooth compactly supported function $\varphi$, either we have $\varphi \le 1/2$ everywhere (case 1) or we don't (case 2). Show that in either case, we have $\|\varphi - 1\|_{W^{1,1}} \ge \frac{1}{2}$. This rules out the possibility of having a sequence of such functions converging to $1$ in $W^{1,1}$ norm.
For case 2, it will help to note that by the fundamental theorem of calculus, for every smooth function $\psi$ and every $x,y$, we have $$\left| \psi(y) - \psi(x) \right| = \left| \int_x^y \psi'(t)\,dt\right| \le \int_x^y |\psi'(t)|\,dt \le \int_{-1}^1 |\psi'(t)|\,dt.$$
Indeed, a variant of this argument can be used to show that a sequence of smooth functions converging in $W^{1,1}((-1,1))$ norm must also converge uniformly. Clearly a sequence of compactly supported functions cannot converge uniformly to 1.