A Banach space is a normed vector space, where a Cauchy sequence converges {{WRT}} the supplied norm. To prove the theorem, we need to show that a Cauchy sequence in $\struct {\map \CC I, \norm {\,\cdot\,}_\infty}$ converges. We take a Cauchy sequence $\sequence {x_n}_{n \mathop \in \N}$ in $\struct {\map \CC I, \norm {\,\cdot\,}_\infty}$. Then we fix $t \in I$ and show, that a real Cauchy sequence $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R, \size {\, \cdot \,}}$ with the limit $\map x t$. Then we prove the continuity of $\map x t$. Finally, we show that $\sequence {x_n}_{n \in \N}$ converges in $\struct {\map \CC I, \norm {\,\cdot\,}_\infty}$ with the limit $\map x t$. === $\sequence {\map {x_n} t}_{n \mathop \in \N}$ is a Cauchy sequence === Let $\sequence{x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $\map \CC I$:  :$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \forall n, m > N : \norm{x_n − x_m}_\infty < \epsilon$ Fix $t \in I$. Then: {{begin-eqn}} {{eqn | l = \size {\map {x_n} t - \map {x_m} t}       | o = \le       | r = \max_{\tau \mathop \in I} \size {\map {x_n} \tau - \map {x_m} \tau} }} {{eqn | r = \norm {x_n - x_m}_\infty       | c = {{Defof|Supremum Norm}} }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} Hence $\sequence{\map {x_n} t}_{n \mathop \in \N}$ is a Cauchy sequence in $\struct {\R, \size {\, \cdot \,}}$. {{qed|lemma}} === $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R,\size {\, \cdot \,}}$ === From Real Number Line is Complete Metric Space, $\R$ is a complete metric space. Therefore, $\sequence{\map {x_n} t}_{n \mathop \in \N}$ is convergent. Denote the limit by $\map x t : I \to \R$: :$\displaystyle \lim_{n \mathop \to \infty} \map {x_n} t = \map x t$ {{qed|lemma}} === $\map x t$ is continuous === Choose $N$ such that:  :$\forall n, m > N : \norm{x_n - x_m} < \dfrac \epsilon 3$ Let $\tau \in I$. Then $\forall n > N$ and $m = N + 1 > N$: {{begin-eqn}} {{eqn | l = \size {\map {x_n} \tau - \map {x_{N + 1} } \tau }       | o = \le       | r = \norm {x_n - x_{N + 1} }_\infty }} {{eqn | o = <       | r = \frac \epsilon 3 }} {{end-eqn}} Take the limit $n \to \infty$: {{begin-eqn}} {{eqn | r = \lim_{n \mathop \to \infty} \size {\map {x_n} \tau - \map {x_{N + 1} } \tau}        | l = \size {\map x \tau - \map {x_{N + 1} } \tau}        | c = $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R,\size {\, \cdot \,} }$ }} {{eqn | o = <        | r = \frac \epsilon 3 }} {{end-eqn}} By assumption, $\map {x_{N + 1} } t \in \map \CC I$. Then: :$\forall t, \tau \in I : \exists \delta > 0: \size {\tau - t} < \delta \implies \size {\map {x_{N + 1} } t - \map {x_{N + 1} } \tau} < \dfrac \epsilon 3$ Thus: {{begin-eqn}} {{eqn | l = \size {\map x \tau - \map x t}        | r = \size {\map x \tau - \map {x_{N + 1} } \tau + \map {x_{N + 1} } \tau - \map {x_{N + 1} } t + \map {x_{N + 1} } t - \map x t}  }} {{eqn | o = <       | r = \size {\map x \tau - \map {x_{N + 1} } \tau} + \size {\map {x_{N + 1} } \tau - \map {x_{N + 1} } t} + \size {\map {x_{N + 1} } t - \map x t}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = <       | r = \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3 }} {{eqn | r = \epsilon}} {{end-eqn}} Hence, $\map x t$ is continuous in $I$. {{qed|lemma}} === $\sequence {x_n}_{n \mathop \in \N}$ converges to $x$=== Let $\epsilon > 0$. Choose $N$ such that: :$\forall n, m > N : \norm {x_n - x_m}_\infty < \epsilon$ Fix $n > N$. Let $t \in I$. Then, $\forall m > N$: {{begin-eqn}} {{eqn | l = \size {\map {x_n} t - \map {x_m} t}       | o = \le       | r = \norm {x_n - x_m}_\infty }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} Take the limit $m \to \infty$: {{begin-eqn}} {{eqn | l = \lim_{m \mathop \to \infty} \size {\map {x_n} t - \map {x_m} t}       | r = \size {\map {x_n} t - \map x t}       | c = $\sequence {\map {x_n} t}_{n \mathop \in \N}$ converges in $\struct {\R,\size {\, \cdot \,} }$ }} {{eqn | o = \le        | r = \max_{t \mathop \in I} \size {\map {x_n} t - \map x t} }} {{eqn | r = \norm {x_n - x}_\infty       | c = {{Defof|Supremum Norm}} }} {{eqn | o = <       | r = \epsilon }} {{end-eqn}} This holds for every $n > N$. Repeat the argument for all $\epsilon > 0$: :$\forall \epsilon > 0 : \exists N \in \N : \forall n \in \N : \forall n > N : \norm {x_n - x}_\infty < \epsilon$. Therefore, $x_n$ converges to $x$ in $\struct {\map \CC I, \norm {\,\cdot\,}_\infty}$: :$\displaystyle \lim_{x \mathop \to \infty} x_n = x$ {{qed}}