We define the Sobolev space $W_p^1 (a,b)$ as:
$$W_p^1 (a,b) = \left\{ u \in L_p (a,b) \ : \ u \in AC\left[a, b\right], \ u' \in L_p \left[a, b\right] \right\}$$
where $AC\left[a,b\right]$ means "absolutely continuous on $\left[a,b\right]$. We defined absolute continuity as follows:
$u$ is absolutely continuous on $\left[a,b\right]$ $\Leftrightarrow$ $u(x) = u(a) + \int_a^x v(t) dt$ for $x \in (a,b)$
We know, that $W_p^1 (a,b) \subset L_p (a,b)$.
We also know, that if $X$ is a complete space, then $A \subset X$ is complete $\Leftrightarrow$ if $A$ is closed
The space $L_p (a,b)$ is complete. So, in order to prove that $W_p^1 (a,b)$ is also complete, we only have to show that $W_p^1 (a,b)$ is closed.
But how exactly? By using the complement of $W_p^1 (a,b)$? By creating a sequence?
The space $W^{1,p}$ is not closed as a subset of $L^p$. To see this, consider the subspace of all $C^1$ compactly supported functions: $$ C_c^1 \subseteq W^{1,p} \subseteq L^p. $$ You're probably familiar with the fact that $C_c^1$ is dense in $L^p$. This means that $\overline{C_c^1} = L^p$, so in particular also $\overline{W^{1,p}} = L^p$ (a larger subset has a larger closure).
This argument proves that $W^{1,p}$ with the $L^p$ norm is not a complete space. However, $W^{1,p}$ is complete with its own norm $$ \| u \|_{W^{1,p}} := \| u \|_{L^p}+\| u' \|_{L^p} $$ To prove completeness of $W^{1,p}$, you can indeed use completeness of $L^p$, but in another way - see my answer to your other question.