Let $\Omega$ be an open set in $\mathbb{R}^n$. For $i\ge1$ define
$$K_i=\overline{B(0,i)}\cap\Big(\bigcup_{y\in\Omega^c}B\Big(y,\frac{1}{i}\Big)\Big)^c.$$
One can show that $K_i$ is compact, $K_i\subset K_{i+1}$ and $\Omega=\bigcup_{i=1}^{\infty}K_i$. Choose $i_0\in\mathbb{N}$ such that $K_{i_0}$ is not empty. Then, it is easy to verify that
$$p_i(f)=\sup_{x\in K_i}|f(x)|,\qquad\text{for}\qquad f\in C(\Omega), \qquad i\ge i_0$$
is a semi-norm on $C(\Omega)$. Using this, it can also be shown that
$$d(f,g)=\sum_{i=i_0}^\infty \frac{2^{-i}p_i(f-g)}{1+p_i(f-g)},\qquad\text{for}\qquad f,g\in C(\Omega)$$
is a metric on $C(\Omega)$.
Is $C(\Omega)$ complete with the metric $d(f,g)$?
My approach was to show that this metric is equivalent to the usual supremum metric which we may define as
$$p(f-g)=\sup_{x\in \Omega}|f(x)-g(x)|,\qquad\text{for}\qquad f,g\in C(\Omega).$$
So, I tried to find lower and upper bounds for $d(f,g)$ in terms of $p(f-g)$. I started with the fact that since $K_i\subset \Omega$ then
$$p_i(f-g)\le p(f-g).$$
Then it can be concluded that
$$\frac{p_i(f-g)}{1+p_i(f-g)}\le\frac{p(f-g)}{1+p(f-g)}.$$
With this in mind, I could find an upper bound as follows
\begin{align*} d(f,g)&=\sum_{i=i_0}^\infty \frac{2^{-i}p_i(f-g)}{1+p_i(f-g)} \\ &\le\frac{p(f-g)}{1+p(f-g)}\sum_{i=i_0}^{\infty}2^{-i}=\frac{p(f-g)}{1+p(f-g)}2^{-i_0+1} \\ &\le \frac{p(f-g)}{1+p(f-g)} \\ & \lt p(f-g). \end{align*}
However, I couldn't find a way to make a lower bound. Can someone help me with this? Also, any other way to show that the metric space $(C(\Omega),d)$ is complete is welcome.
As suggested by Daniel in the comments, let $(f_n)$ be a Cauchy sequence in $(C(\Omega),d)$. Then, by definition, we have
$$\forall \epsilon>0,\quad\exists N\in\mathbb{N},\quad\forall n,m,\quad n,m\ge N\quad \implies d(f,g)=\sum_{i=i_0}^\infty \frac{2^{-i}p_i(f_n-f_m)}{1+p_i(f_n-f_m)}\lt\epsilon.$$
This in turn results in
$$\forall \epsilon>0,\quad\exists N\in\mathbb{N},\quad\forall n,m,\quad n,m\ge N\quad \implies \tilde d_i(f_n,f_m)=\frac{2^{-i}p_i(f_n-f_m)}{1+p_i(f_n-f_m)}\lt\epsilon,\quad i=i_0,\dots$$
since
$$\frac{2^{-i}p_i(f_n-f_m)}{1+p_i(f_n-g_m)}\le \sum_{i=i_0}^\infty \frac{2^{-i}p_i(f_n-f_m)}{1+p_i(f_n-f_m)}\lt\epsilon,\quad i=i_0,\dots$$
According to the definition of $p_i$, one can replace $f_n$ and $f_m$ with $f_n|_{K_i}$ and $f_m|_{K_i}$ since $\sup_{x\in K_i}|f(x)|=\sup_{x\in K_i}|f|_{K_i}(x)|$ or equivalently $p_i(f)=p_i(f|_{K_i})$. This means that $(f_n|_{K_i})$ is Cauchy in $(C(K_i),\tilde d_i)$. According to this post, being Cauchy and convergent is equivalent for $\tilde d_i$ and $p_i$. Consequently, $(f_n|_{K_i})$ is Cauchy in $(C(K_i),p_i)$. But $(C(K_i),p_i)$ is complete so $(f_n|_{K_i})$ converges to some $g_i$ in $(C(K_i),p_i)$. Now, we introduce the following candidate $g:\Omega\to\mathbb{R}$
$$g|_{K_i}:=g_i=\lim_{n\to\infty} f_n|_{K_i}\quad\text{in}\quad(C(K_i),p_i),\quad i=i_0,\dots$$
$g$ is well defined since for every $i\ge j$, it is readily verified that $g|_{K_i}|_{K_j}=g|_{K_j}$ (See the comment below this answer for more details on this). Note that $g|_{K_i}$ is continuous over $K_i$ as it is the uniform limit of $f_n|_{K_i}$. This implies that $g\in C(\Omega)$ because for every $x\in\Omega$ there is an $i$ such that $x\in \text{int} K_i$. Now, we are ready to go on. As the convergence with respect to $p_i$ implies convergence with respect to $\tilde d_i$, $(f_n|_{K_i})$ converges to $g|_{K_i}$ in $(C(K_i),\tilde d_i)$. This essentially means that
$$\forall \tilde\epsilon>0,\quad\exists \tilde N\in\mathbb{N},\quad\forall n,\quad n\ge \tilde N\quad \implies \tilde d_i(f_n,g)=\frac{2^{-i}p_i(f_n-g)}{1+p_i(f_n-g)}\lt\tilde\epsilon,\quad i=i_0,\dots$$
and we want to show that
$$\forall \epsilon>0,\quad\exists N\in\mathbb{N},\quad\forall n,\quad n\ge N\quad \implies d(f,g)=\sum_{i=i_0}^\infty \frac{2^{-i}p_i(f_n-g)}{1+p_i(f_n-g)}\lt\epsilon.$$
and this can be done since
\begin{align*} \sum_{i=i_0}^\infty \frac{2^{-i}p_i(f_n-g)}{1+p_i(f_n-g)}&= \sum_{i=i_0}^{j_0} \frac{2^{-i}p_i(f_n-g)}{1+p_i(f_n-g)} + \sum_{i=j_0+1}^\infty \frac{2^{-i}p_i(f_n-g)}{1+p_i(f_n-g)} \\ &\le \sum_{i=i_0}^{j_0} \frac{2^{-i}p_i(f_n-g)}{1+p_i(f_n-g)}+ \sum_{i=j_0+1}^\infty 2^{-i} \\ &= \sum_{i=i_0}^{j_0} \frac{2^{-i}p_i(f_n-g)}{1+p_i(f_n-g)}+ 2^{-j_0}. \end{align*}