motivation and geometric intuition of considering the metric $\hat{d}$ when showing open subset of Polish is Polish In this set of descriptive set theory's notes, page $10,$ the author has the following lemma.

Lemma $3.4$ If $X$ is a Polish space (separable completely metrizable) and $U$ is an open subset of $X,$ then $U$ (with the subspace topology) is a Polish space.

In his proof, he let $d$ be a complete metric on $X$ such that $d$ is compatible with the topology on $X$ and $d<1.$
Then he proceed to define
$$\hat{d}(x,y) = d(x,y) + \bigg| \frac{1}{d(x,X\setminus U)} - \frac{1}{d(y,X\setminus U)} \bigg|,$$
where $x,y\in U$ and
$$d(x,X\setminus U) = \inf_{z\in X\setminus U}  d(x,z).$$
Then the author proceed to show that $\hat{d}$ and $d$ generate the same topology and $\hat{d}$ is complete in $U.$

Question: What is a motivation and geometric intuition behind the metric $\hat{d}?$

I notice that Cauchy sequence with respect to $\hat{d}$ must be Cauchy with respect to $d,$ but not the converse.
So I am thinking that $\hat{d}$ 'throw away' some Cauchy sequences with respect to $d$ which do not converge in $U.$
But how do we know we have 'thrown ' all such Cauchy sequences?
Particularly, how does the term
$$\bigg| \frac{1}{d(x,X\setminus U)} - \frac{1}{d(y,X\setminus U)} \bigg|$$
ensure all 'survive' Cauchy sequences do converge in $U?$
 A: (*). The intuition is that for the metric $\hat d$ on $U$ the set $X$ \ $U$ is infinitely far away in this sense:  A $d$-Cauchy sequence of members of $U$ that converges to  a member of $X$ \ $U$ is $\hat d$-unbounded and therefore not a $\hat d$- Cauchy sequence.  But any $d$-Cauchy sequence of members of $U$ that converges to a member $x$ of $U$ is also $\hat d$-convergent to $x$ because $d|_U$ and $\hat d$ generate the same topology on $U$.  (See  (***)  below).
Furthermore a $\hat d$-Cauchy sequence $S$ is also a $d$-Cauchy sequence because $\hat d \geq d$, so $S$ is $d$-convergent to some $x\in X$.  By the above paragraph, $x\in U$.  So $S$ must be $\hat d$-convergent to $x$ because equivalent metrics have the same set of convergent sequences. 
(**). I don't think it is necessary in the Q that $d<1.$
Examine the example with $X=\Bbb R$ and $d(x,y)=|x-y|$ and $U=\Bbb R^+.$
(***). Theorem: If $(U,d)$ and $(V,e)$ are metric spaces and $f:U\to V$ is continuous then $\hat d(x,y)=d(x,y)+e(f(x),f(y))$ is a metric on $U$ equivalent to $d.$ (That is, $d$ and $\hat d$ generate the same topology.) 
This is useful for various constructions, particularly with $Y=\Bbb R$ and $e(c,d)=|c-d|.$  In the Q we use $f(x)=d(x, X$ \ $U)^{-1}.$
(****). The more general result is that if  $Y$ is a subspace of a completely metrizable space $X$ then $Y$ is completely metrizable iff $Y$ is $G_{\delta}$ in $X.$ And this can be extended a little:  (I). If $Y$ is a completely metrizable subspace of the metric space $X$ then $Y$ is $G_{\delta}$ in $X.$ (II). If $X$ is a Hausdorff space  and $F$ is a countable (i.e. finite or countably infinite) family of completely metrizable subspaces of $X$ then $\cap F$ is completely metrizable.  
