$C[0,1]$ is not complete with the metric $d(f,g)$. To show it pick the sequence
$(f_n)_n\subset C[0,1]$ defined by
$$
f_n(x)=\left\{\begin{array}{ll}0,&0\leq x<\frac{1}{2}-\frac{1}{n}\\n(x-\frac{1}{2}+\frac{1}{n}),&\frac{1}{2}-\frac{1}{n}\leq x<\frac{1}{2}\\1,&\frac{1}{2}\leq x\leq1 \end{array}\right.
$$
Draw the first ones before you keep reading! This is a Cauchy sequence as a consequence of the estimation
$$
\int_0^1(f_n-f_m)^2dx\leq\int_{1/2-1/n}^{1/2}f_n^2dx\leq\int_{1/2-1/n}^{1/2}f_ndx=\frac{1}{2n},\hspace{.8cm}n\geq m,
$$
but there are no $f\in C[0,1]$ such that $d(f_n,f)\to0$ because $f_n(x)\to\chi_{[1/2,1]}(x)$ for every $x$ in $[0,1]$.
I saw your question a while ago, but I didn't know how to answer it. As time have passed and nobody answered you, I'll write my approach, which I don't know if it's correct.
Let $\{[a_n,b_n]\}_{n\geq 1}\subset I([0,1])$ be a Cauchy sequence. Then, for every $\varepsilon>0$ exists $n_0\in\mathbb{N}$ such that $d_H([a_n,b_n],[a_m,b_m])<\varepsilon$, for every $n,m>n_0$. Now, for intervals, the following relation is verified:
\begin{equation*}
d_H([a,b],[c,d])=\max\{|a-c|,|b-d|\}.
\end{equation*}
Then, in our case, for every $\varepsilon>0$ exists $n_0\in\mathbb{N}$ such that $\max\{|a_n-a_m|,|b_n-b_m|\}<\varepsilon$, for every $n,m>n_0$. In other words, $|a_n-a_m|<\varepsilon$ and $|b_n-b_m|<\varepsilon$, for every $n,m>n_0$. With this, we observe that $\{a_n\}_{n\geq 1}$ and $\{b_n\}_{n\geq 1}$ are both Cauchy sequences of real numbers. Since $(\mathbb{R},|\cdot|)$ is a complete space, there exist $a,b\in\mathbb{R}$ such that $a_n\rightarrow a$ and $b_n\rightarrow b$.
To end, let's see that $[a_n,b_n]\rightarrow [a,b]$, in the sense of the Hausdorff distance. Using the same property as before we have that
\begin{equation*}
d_H([a_n,b_n],[a,b])=\max\{|a_n-a|,|b_n-b|\}\rightarrow 0,
\end{equation*}
since $a_n\rightarrow a$ and $b_n\rightarrow b$. So, the Cauchy sequence $\{[a_n,b_n]\}_{n\geq 1}$ is a convergent sequence, which finish the proof.
I hope you find this useful and, if you spot any mistake in the reasoning, please comment it.
Best Answer
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.