Help understanding Rudin's proof showing that $C_c(X)$ is dense in $L^p(\mu)$ The proof is from Rudin's "Real and Complex Analysis." It states

For $1\leq p<\infty$, $C_c(X)$ is dense in $L^p(\mu)$

The proof is

Let $S$ be the class of all complex, measurable, simple functions on $X$ such that $\mu(\{x:s(x)\neq 0\})<\infty$. If $s\in S$ and $\epsilon<0$, there exists a $g\in C_c(X)$ such that $g(x)=s(x)$ except on a set of measure $<\epsilon$ and $\vert g\vert\leq\,\parallel s\,\parallel_{\infty}$ (Lusin's Theorem). Hence $$\parallel g-s\parallel_p\leq 2\epsilon^{1/p}\parallel s\,\parallel_{\infty}$$
  Since $S$ is dense in $L^p(\mu)$, this completes the proof.

I have trouble understanding the inequalities $$\vert g\vert\leq\,\parallel s\,\parallel_{\infty}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$$$\parallel g-s\parallel_p\leq 2\epsilon^{1/p}\parallel s\,\parallel_{\infty}\,\,\,\,\,\,\,\,\,\,\,(2)$$
I know that the first inequality is supposed to follow from Lusin's Theorem, which in this case would result in $$\sup_{x\in X}\vert g(x)\vert\leq\sup_{x\in X}\vert s(x)\vert$$
However, I am not sure how to get to $(1)$ from here. Lastly, I don't understand how to get inequality $(2)$.
Added: I forgot to mention that $X$ is supposed to be a locally compact Hausdorff space and that $\mu$ is a measure on a $\sigma$-algebra $\mathfrak M$ in $X$ with the properties of the measure of Riesz's Representation Theorem.

For (1),
$$
|g|\leq\sup|g(x)|\leq\sup|s(x)|=\|s\|_\infty.
$$
For (2),
\begin{align}
\|g-s\|_p^p&=\int_X|g-s|^p\,d\mu=\int_{g\ne s}|g-s|^p\,d\mu
\leq\int_{g\ne s}(|g|+|s|)^p\,d\mu
\leq\int_{g\ne s}(2\|s\|_\infty)^p\,d\mu\\
&\leq2^p\|s\|_\infty^p\,\epsilon.
\end{align}
