$W_0^{1,\:p}(\Lambda)$ is dense in $L^2(\Lambda)$

Question

Let

• $d\in\mathbb N$
• $\lambda$ be the Lebesgue measure on $\mathbb R^d$
• $\Lambda\subseteq\mathbb R^d$ be open with $\lambda(\Lambda)<\infty$
• $p\ge 2$
• $W^1(\Lambda)$ denote the set of weakly differentiable functions $\Lambda\to\mathbb R$, $$W^{1,\:p}(\Lambda):=\left\{u\in W^1(\Lambda)\cap L^p(\Lambda):\nabla u\in L^p(\Lambda,\mathbb R^d)\right\}$$ be equipped with $$\left\|u\right\|_{W^{1,\:p}(\Lambda)}:=\left(\left\|u\right\|_{L^p(\Lambda)}+\left\|\nabla u\right\|_{L^p(\Lambda,\:\mathbb R^d)}\right)^{\frac 1p}\;\;\;\text{for }u\in W_0^{1,\:p}(\Lambda)$$ and \begin{equation} \begin{split} W_0^{1,\:p}(\Lambda)&:=\left\{u\in W^{1,\:p}(\Lambda)\mid\exists(\varphi_n)_{n\in\mathbb N}\subseteq C_c^\infty(\Lambda):\left\|\varphi_n-u\right\|_{W^{1,\:p}(\Lambda)}\stackrel{n\to\infty}\to 0\right\}\\ &=\overline{C_c^\infty(\Lambda)}^{\left\|\;\cdot\;\right\|_{W_0^{1,\:p}}} \end{split} \end{equation}

I'm trying to understand why

1. $W_0^{1,\:p}(\Lambda)$ is dense in $L^2(\Lambda)$
2. On $W_0^{1,\:p}(\Lambda)$, the $\left\|\;\cdot\;\right\|_{W^{1,\:p}(\Lambda)}$-topology $\tau_1$ is finer than the $\left\|\;\cdot\;\right\|_{L^2(\Lambda)}$-topology $\tau_2$

Since $\lambda(\Lambda)<\infty$, it's obvious that $C_c^\infty(\Lambda)\subseteq L^p(\Lambda)$. Thus, since $C_c^\infty(\Lambda)$ is dense in $L^2(\Lambda)$, it's clear that $L^p(\Lambda)$ is dense in $L^2(\Lambda)$, if I'm not terribly wrong.

However, I don't know how we can conclude (1.) and (2.). So, how can we prove these statements?