Show that $C([0,1],\mathbb{R}^{2})$ is complete. I need to show that $C([0,1],\mathbb{R}^{2})$ is complete.
$C([0,1],\mathbb{R}^{2})$ are continuous functions $f:[0,1]\to\mathbb{R}^{2}$ and $\;d(f,g)=\sup_{x\in[0,1]}\left\Vert f(x)-g(x)\right\Vert _{2}\;$ where $\left\Vert \cdot\right\Vert _{2}$ is the standard euclidean distance over $\mathbb{R}^{2}$.
My problem is, is that I don't know how to even approach this. I know that it is sufficient to show that for each Cauchy sequence we can find a convergent sub-sequence, but it seems too abstract to follow that path.
What am I missing?

Outline of proof:
Let $(f_n)$ be Cauchy in $X=C([0,1],\mathbb{R}^2).$ Then $(f_n(x))$ is Cauchy in $\mathbb{R}^2$ for each $x\in[0,1].$ Since $\mathbb{R}^2$ is complete, it follows that $(f_n)$ is pointwise convergent, to say $f.$
Now show that $f$ is continuous and $f_n\to f$ uniformly.
Edit: Let $\epsilon >0,$ then there exists $n_0 \in \mathbb{N}$ such that $d(f_n,f_m)<\frac{\epsilon}{2}$ for all $n,m\geq N.$ Now for each $x,f_n(x) \to f(x),$ so choose $n_x\geq N$ such that $$\|f_{n_x}(x)-f(x)\|_2<\frac{\epsilon}{2}.$$ We have \begin{align*}\|f_n(x)-f(x)\|_2&\leq\|f_n(x)-f_{n_x}(x)\|_2+\|f_{n_x}-f(x)\|_2\\&< d(f_n,f_{n_x})+\frac{\epsilon}2\\&<\epsilon\end{align*} for all $n\geq N,x \in [0,1].$ This implies $d(f_n,f)\leq \epsilon$ for all $n \geq N$ and hence $f_n \to f$ uniformly.
