Consider the set $X = \{f: I \to \mathbb{R}^2$:$f$ is continuous} (here $I = [0,1]$, prove that $(X,d_\infty)$ is complete where $d_\infty(f,g) =\sup_{x \in I}|(f(x)-g(x)|$. The hint we are given is that this proof doesn't differ much from the case when $X$ consists of real valued continuous functions $f:[0,1] \to \mathbb{R}$. With that is mind, this is my attempt of the proof,
Suppose ${f_n}$ is a Cauchy sequence in $d_\infty$, then given $\epsilon > 0$ there is an $N(\epsilon)$ such that $|f_n(x) - f_m (x)|<\epsilon$ for all $x \in I$, then the sequence ${f_n (x)}$ is a Cauchy sequence in $\mathbb{R}^2$ which converges to a value $f(x)$ by the completeness of $\mathbb{R}^2$ under $d_\infty$. Since $|f_n(x) - f_m (x)|<\epsilon$ for all $x \in I, n,m \geq N(\epsilon), \lim_{m \to \infty}|f_n(x) - f_m(x)| = |f_n(x) - f(x)| < \epsilon$ for all $x \in I$ for $n \geq N(\epsilon)$. So $\sup|f_n(x)- f(x)| < \epsilon$ for $n \geq N(\epsilon)$ and since $f$ is continuous it converges to some point in $X$, hence $(X,d_\infty)$ is complete.
Now this proof feels very shaky to me, especially as it relies heavily on the italicised sentence which although I know is true I don't know if I can assume it, can I get some feedback/help? Thanks