is the function $\rho$ a pseudometric? Let $\Im=\left\{\Im_{n}\right\}_{n\in \mathbb{N}}$ be a sequence of open covers of a topological space $(X,\tau)$.
We define a function $\delta:X\times X \rightarrow \mathbb{R}$ as follows
If $(x,y)\in X\times X$, let
1) $\delta(x,y)=2$ if $\left\{x,y\right\}$ does not refine $\Im_{1}$.
2) $\delta(x,y)=\frac{1}{2^{n-1}}$ if $\left\{x,y\right\}$ refines $\Im_{n}$ but does not refine $\Im_{m}$ for all $m>n$.
3) $\delta(x,y)=0$ if $\left\{x,y\right\}$ refines $\Im_{n}$ for all $n\in \mathbb{N}$.
I already proved
a) $\delta(x,x)=0$
b) $\delta(x,y)=\delta(y,x)$
c) $\delta(x,y)\leq 2\cdot \max\left\{\delta(x,z),\delta(y,z)\right\}$.
But I want to obtain a pseudometric, so the function $\rho:X\times X \rightarrow \mathbb{R}$ is defined as follows.
if $(x,y)\in X\times X$, let
$\rho(x,y)=\inf\left\{\delta(x,x_{1})+\delta(x_{1},x_{2})+\cdots + \delta(x_{n},y)\right\}$
where the infimum is taken over all finite subsets $\left\{x_{1},\ldots ,x_{n}\right\}$ of $X$.
The problem is here I can't prove that $\rho$ is a pseudometric; I can't prove that if $x=y$ then $\rho(x,x)=0$ and $\rho(x,y)\leq \rho(x,z) + \rho(z,y)$ for all $x,y,x \in X$.
And also I can't prove that for all $(x,y)\in X\times X$
$\rho(x,y)\leq \delta(x,y)$ 
$\delta(x,y)\leq 2^{2}\rho(x,y)$.
if someone has a hint, please share it with me.
 A: If you allow arbitrary sequences of open covers, your function $\delta$ isn’t necessarily defined: you might have $\{x,y\}$ refining $\mathscr{T}_n$ if and only if $n$ is even, in which case none of your three conditions holds. If you assume that $\mathscr{T}_{n+1}$ refines $\mathscr{T}_n$ for each $n$, you avoid this problem, because now for each $x,y\in X$ either


*

*$\{x,y\}$ does not refine $\mathscr{T}_0$, or  

*$\{x,y\}$ refines $\mathscr{T}_n$ for each $n\in\Bbb N$, or  

*there is a unique $n\in\Bbb N$ such that $\{x,y\}$ refines $\mathscr{T}_m$ if and only if $m\le n$.


But this still isn’t enough to ensure that $\delta(x,y)\le 2\max\{\delta(x,z),\delta(z,y)\}$. The simplest way to ensure that is to require that for each $n\in\Bbb N$ and $U,V\in\mathscr{T}_{n+1}$ such that $U\cap V\ne\varnothing$ there is a $W\in\mathscr{T}_n$ such that $U\cup V\subseteq W$; in this case $\mathscr{T}_{n+1}$ is said to be a regular refinement of $\mathscr{T}_n$.

Added after seeing later comments above: I expect that either your $2$-refinement is what I call a regular refinement, or it’s easily seen that a $2$-refinement is a regular refinement.

You want to define $\rho(x,y)$ over all finite paths from $x$ to $y$ including the one-step path directly from $x$ to $y$:
$$\rho(x,y)=\min\left\{\sum_{k=0}^{n-1}\delta(x_k,x_{k+1}):n\in\Bbb Z^+\text{ and }x_0=x\text{ and }x_n=y\text{ and all }x_k\in X\right\}\;.\tag{1}$$
Now one of the sums in $(1)$ is the unique sum with $n=1$,
$$\sum_{k=0}^{1-0}\delta(x_0,x_{0+1})=\delta(x,y)\;,$$
and it should be immediately clear that $\rho(x,y)\le\delta(x,y)$: the minimum of a set of real numbers cannot be larger than any of those numbers. This immediately implies that $\rho(x,x)=0$ for all $x\in X$, since $0\le\rho(x,x)\le\delta(x,x)=0$.
For the triangle inequality, suppose that there are some $x,y,z\in X$ such that
$$\rho(x,y)>\rho(x,z)+\rho(z,y)\;.$$
Show that this implies that there are $m,n\in\Bbb Z^+$ and points $x_k\in X$ for $k=0,\ldots,m+n$ such that $x_0=x$, $x_m=z$, $x_{m+n}=y$, and
$$\sum_{k=0}^{m+n-1}\delta(x_k,x_{k+1})<\rho(x,y)$$
and get an immediate contradiction.
For the final inequality, prove by induction on $n\ge 2$ that for every path $x=x_0,x_1,\ldots,x_n=y$ the following condition holds:
$$\delta(x,y)\le 2\delta(x,x_1)+4\sum_{k=1}^{n-2}\delta(x_k,x_{k+1})+2\delta(x_{n-1},y)\;.\tag{2}$$
For $n=2$ this is just $\delta(x,y)\le 2\delta(x,x_1)+2\delta(x_1,y)$, which is certainly true, since
$$\begin{align*}
\delta(x,y)&\le 2\max\{\delta(x,x_1),\delta(x_1,y)\}\\
&=\max\{2\delta(x,x_1),2\delta(x_1,y)\}\\
&\le 2\delta(x,x_1)+2\delta(x_1,y)\;.
\end{align*}$$
Suppose that $n\ge 3$, and $(2)$ holds for all $m$ such that $2\le m<n$. Let $x=x_0,x_1,\ldots,x_n=y$ be a path from $x$ to $y$. For $k=1,\ldots,n$ we have
$$\delta(x,y)\le 2\max\{\delta(x,x_k),\delta(x_k,y)\}\;,$$
so either $\delta(x,y)\le 2\delta(x,x_k)$ or $\delta(x,y)\le 2\delta(x_k,y)$. Clearly $\delta(x,y)=\delta(x,x_n)\le 2\delta(x,x_n)$, so there is at least one $k\ge 1$ such that $\delta(x,y)\le 2\delta(x,x_k)$; let $\ell$ be the least such $k$.
If $\ell=1$, then
$$\delta(x,y)\le 2\delta(x,x_1)\le 2\delta(x,x_1)+4\sum_{k=1}^{n-2}\delta(x_k,x_{k+1})+2\delta(x_{n-1},y)\;,$$
so $(2)$ holds. If $\ell>1$, then $\delta(x,y)\le 2\delta(x_{\ell-1},y)$ and $\delta(x,y)\le 2\delta(x,x_\ell)$, so
$$\delta(x,y)=\frac12\delta(x,y)+\frac12\delta(x,y)\le\delta(x,x_\ell)+\delta(x_{\ell-1},y)\;.$$
Now apply the induction hypothesis to the paths $x=x_0,\ldots,x_\ell$ and $x_{\ell-1},\ldots,x_n=y$:
$$\delta(x,x_\ell)\le 2\delta(x,x_1)+4\sum_{k=1}^{\ell-2}\delta(x_k,x_{k+1})+2\delta(x_{\ell-1},x_\ell)\tag{3}$$
and
$$\delta(x_{\ell-1},y)\le 2\delta(x_{\ell-1},x_\ell)+4\sum_{k=\ell}^{n-1}\delta(x_k,x_{k+1})+2\delta(x_{n-1},y)\;,\tag{4}$$
and adding $(3)$ and $(4)$ yields $(2)$.
Finally, suppose that there are $x,y\in X$ such that $\delta(x,y)>4\rho(x,y)$. Then there are an $n\in\Bbb Z^+$ and a path $x=x_0,x_1,\ldots,x_n=y$ such that
$$\begin{align*}
\delta(x,y)&>4\sum_{k=0}^{n-1}\delta(x_k,x_{k+1})\\
&\ge 2\delta(x,x_1)+4\sum_{k=1}^{n-2}\delta(x_k,x_{k+1})+2\delta(x_{n-1},y)\\
&\ge \delta(x,y)\;,\tag{by (2)}
\end{align*}$$
which is absurd. Thus, $\delta(x,y)\le 4\rho(x,y)$ for all $x,y\in X$.
