Use the $\epsilon-\delta$ definition to prove that $\lim_{x\rightarrow c}f(x)=L$ if and only if $\lim_{x\rightarrow c}|f(x)-L|=0$
Let $\epsilon >0.$ Then since we know that $\lim_{x\rightarrow c}f(x)=L$ there exists a $\delta >0$ such that $0<|x-c|< \delta$ implies that $|f(x)-L|< \epsilon$. Then by the triangle inequality it follows that $||f(x)|-|L|| \leq |f(x)-L|< \epsilon.$ Thus $\lim_{x\rightarrow c}|f(x)-L|=0$. Now if $\lim_{x\rightarrow c}|f(x)-L|=0$ then there exists a $\delta >0$ such that $0<|x-c|< \delta$ implies that $||f(x)-L||< \epsilon$. I'm not sure how to get $|f(x)-L|\leq ||f(x)-L||$
The first part is not correct, writing $||f(x)|-|L||<\epsilon$ is to conclude that $\lim_{x\rightarrow c}|f(x)|=|L|$, but this is not part of the question.
Rather, you may write $||f(x)-L|-0|=|f(x)-L|<\epsilon$, so $||f(x)-L|-0|<\epsilon$, so $\lim_{x\rightarrow c}|f(x)-L|=0$.