Proving a contraction mapping is a Cauchy sequence

You're on the right track. Note that since $0<L<1$, we have $$ L^m+L^{m+1}+\dots+L^{m+n-1}\leq \sum_{k=m}^{\infty}L^k=\frac{L^m}{1-L} $$ which can be made as small as we like by choosing $m$ large enough.

You are very close to a complete proof. All you need is the "final step".

Here is your proof, completed with the "final step" (in details).

Let $\phi(x):[a,b]\rightarrow [a,b]$ be a continuous function. Show that if $\phi(x)$ is a contraction mapping on $[a,b]$ then the sequence $\{x^{(k)}\}$ defined by $x^{(k+1)} = \phi(x^{(k)})$ is a Cauchy sequence.

Proof - Since $\phi(x)$ is a contraction mapping we have $$|x^{(k+1)} - x^{(k)}| = |\phi(x^{(k)}) - \phi(x^{(k-1)})|\leq L|x^{(k)} - x^{(k-1)}|$$ where $0\leq L< 1$. It folows by induction (that is, applying this idea repeatedly) we get $$|x^{(k+1)} - x^{(k)}|\leq L^k|x^{(1)} - x^{(0)}|$$ Now consider the term that must be bounded in order to be a Cauchy sequence \begin{align*} |x^{(m)} - x^{(m+n)}| &= |(x^{(m)} - x^{(m+1)}) + (x^{(m+1)} - x^{(m+2)}) + \ldots + (x^{(m+n-1)} - x^{(m+n)})|\\ &\leq |(x^{(m)} - x^{(m+1)})| + |(x^{(m+1)} - x^{(m+2)})| + \ldots + |(x^{(m+n-1)} - x^{(m+n)})|\\ &\leq (L^m + L^{m+1} + \ldots + L^{m+n-1})|x^{(1)} - x^{(0)}|= \\ & =\left( \sum_{k=m}^{m+n-1}L^k \right) |x^{(1)} - x^{(0)}| \leq \left( \sum_{k=m}^{\infty}L^k \right) |x^{(1)} - x^{(0)}| = \\ & = \frac{L^m}{1-L}|x^{(1)} - x^{(0)}| \end{align*}

Given $\varepsilon >0$, since $0\leq L <1$, there is $M\in \mathbb{N}$ such that for all $m>M$ and all $n \in \mathbb{N}$, $$|x^{(m)} - x^{(m+n)}| \leq \frac{L^m}{1-L}|x^{(1)} - x^{(0)}| \leq \varepsilon$$

So the sequence $\{x^{(k)}\}$ is a Cauchy sequence.