How to resolve the apparent paradox resulting from two different proofs?

Definition of Open Ball
Let $(X, d)$ be a metric space and let $r\in\mathbb{R}^+$. Then the set,
$B_d(x, r) := \{y \in X : d(x, y) < r\}$
will be said to be the open ball of radius $r$ centered at $x$ in the metric space $(X, d)$.

Definition of Open Set

Let $(X,d)$ be a metric space and $U\subseteq X$. Then $U$ will be said to be $d$-open in $X$ if for all $x\in U$, there exists $r>0$ such that $B_d(x,r)\subseteq U$.

Problem. Let $(X,d)$ be a metric space and $x,y\in X$. Let $B_d(x,r)$ and $B_d(y,s)$ be two open balls in $X$. Show that $B_d(x,r_x)\cap B_d(y,r_y)$ is open in $X$.

First Proof.

If $B_d(x,r_x)\cap B_d(y,r_y)=\emptyset$ then we have nothing to prove since $\emptyset$ is open in $X$.

Otherwise, $B_d(x,r_x)\cap B_d(y,r_y)\ne\emptyset$ and let $z\in B_d(x,r_x)\cap B_d(y,r_y)$. Choose $r_z=\min\{r_x-d(x,z),r_y-d(y,z)\}$. Now consider the open ball $B_d(z,r_z)$. Let $u\in B_d(z,r_z)$. Then we have, $$d(x,u)\le d(x,z)+d(z,u)<r_x$$which shows that $u\in B_d(x,r_x)$. Also, $$d(y,u)\le d(y,z)+d(z,u)<r_y$$which shows that $u\in B_d(y,r_y)$. Consequently we have, $u\in B_d(x,r_x)\cap B_d(y,r_y)$. So, we conclude that $B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$ and we are done.

Second Proof.

If $B_d(x,r_x)\cap B_d(y,r_y)=\emptyset$ then we have nothing to prove since $\emptyset$ is open in $X$.

Otherwise, $B_d(x,r_x)\cap B_d(y,r_y)\ne\emptyset$ and let $z\in B_d(x,r_x)\cap B_d(y,r_y)$. If we can show that there exists $r_z\in\mathbb{R}^+$ such that $B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$ then we are done.

So, let us assume that such $r_z$ exists and let $u\in B_d(z,r_z)$. Then, $$d(x,u)\le d(x,z)+d(z,u)<d(x,z)+r_z$$If we want that $d(x,z)+r_z\le r_x$
Then we choose $r_z$ to be the infimum of the set, $$R:=\{r_v\in\mathbb{R}:(\exists v\in B_d(x,r_x)\setminus\{x\})[r_v=r_x-d(x,v)]\}$$ provided $R\ne \emptyset$. Such infimum exists since $R$ is bounded below by $0$. Since $r_z=\inf R$ we conclude that $r_z\le r_x-d(x,v)$ for all $v\in B_d(x,r_x)$. Hence in particular for $v=z$ we have $r_z\le r_x-d(x,z)$. We conclude that if such $r_z$ exists then $r_z\le r_x-d(x,z)$.

In a similar manner we will be able to show that $r_z\le r_y-d(y,z)$. Consequently we get, $$r_z\le \min \{r_x-d(x,z),r_y-d(y,z)\}$$i.e., if such an $r_z$ exists then $r_z\le \min \{r_x-d(x,z),r_y-d(y,z)\}$.

Now let us choose $r_z\in (0,\min \{r_x-d(x,z),r_y-d(y,z)\}]$. Now consider the open ball $B_d(z,r_z)$. Let $u\in B_d(z,r_z)$. Then we have, $$d(x,u)\le d(x,z)+d(z,u)<r_x$$which shows that $u\in B_d(x,r_x)$. Also, $$d(y,u)\le d(y,z)+d(z,u)<r_y$$which shows that $u\in B_d(y,r_y)$. Consequently we have, $u\in B_d(x,r_x)\cap B_d(y,r_y)$. So, we conclude that $B_d(z,r_z)\subseteq B_d(x,r_x)\cap B_d(y,r_y)$ and we are done.

There is an important difference in the two proofs. In the first proof we don’t need to consider the case when $r_z=0$ but in the second proof we need to consider that (although I haven’t done that, and I think that it is a flaw) because $\inf R$ can be $0$. But then we need to define what do we mean by an open ball of radius $0$. Probably that will be $\emptyset$ as has been suggested here. The question is,

Is it “wrong” to define an open ball in a metric space $(X,d)$ as has been defined above? If not then how can one resolve the apparent paradox that results from the second proof?

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In the second proof, everything between “So, let us assume…” and “…then $r_z\le \min \{r_x-d(x,z),r_y-d(y,z)\}$.” is not really part of the proof (you may notice that nothing proven in that section is ever used to make any logical deduction in the remainder of the proof). Rather, it is motivation for the following line of the proof: “Now let us choose $r_z\in (0,\min \{r_x-d(x,z),r_y-d(y,z)\}]$”. The preceding section attempts to show why this is a wise choice of $r_z$, but you could still make this choice of $r_z$ even if you didn’t provide any reason to think it was a wise choice (which is what is done in the first proof).

In particular, you don’t care if $\inf R$ could be $0$, because nothing about $R$ is actually necessary for the validity of the second proof.

[In fact, the motivation provided by the discussion about $R$ is not actually good motivation. In most examples, $\inf R$ actually will be $0$, and so it was wrong to think that $r_z$ was forced to be $\inf R$. Indeed, it is unclear to me why you defined $R$ at all, or why you think $\inf R$ is a good choice for $r_z$.]

The second proof is flawed:
The defintion is R is incorrect
$$R:=\{r_v\in\mathbb{R}:(\forall v\in B_d(x,r_x))[r_v=r_x-d(x,v)]\}$$. I guess it might be
$$R:=\{r_v\in\mathbb{R}:(\exists v\in B_d(x,r_x))[r_v=r_x-d(x,v)]\}$$

Also, the infimum of R could be zero. A ball with radius 0 is not open ball. The mistake is the set R is defined too broad so it is useless in the proof.