Euclidean metric on $\mathbb{R^{2}}$ generates the standard (product) topology on $\mathbb{R^{2}}$ 
Below is the definition of product topology.
Let $X$ and $Y$ be sets. The product of $X$ and $Y$ , denoted $X$ and $Y$ is the set of ordered pairs given by 
$$X \times Y := \{(x,y) \vert x \in X, y \in Y\}. ~~$$
I can let $U$ be an open set in $\mathbb{R}^{2}$, then let $x \in U$, then $x$ is an interior point of $U$, so $\exists r > 0$, such that $B_r(x) = \{y \vert d(x,y) < r \} \subseteq U$. Therefore, $d(x,y)$ is the generated standard topology on $\mathbb{R}^{2}$. This is how I thought I will do the problem but cant get it properly. 
Can someone help me on this. Any help will be greatly appreciated. 
 A: There are two definitions of the product topology on $\Bbb R^2$, which are equivalent: one is the topology generated by the base
$$\mathcal{B}=\{U \times V: U,V \subseteq \Bbb R \text{ open }\}$$
and another is more abstractly as "the smallest topology that makes both maps $\pi_1,\pi_2: \Bbb R^2 \to \Bbb R$, defined by $\pi_1(x,y)=x$ and $\pi_2(x,y)=y$, continuous"
I'll use the former, more concrete, view.
Let $O$ be product open in $\Bbb R^2$, and we have to show that $O$ is $d$-open, or equivalently that each $(x_1,x_2) \in O$ is an interior point of $O$ w.r.t. $d$.
So let $(x_1,x_2) \in O$ and by the (first) definition of the product topology we can find $O_1,O_2$ open in $\Bbb R$ such that $x_1 \in O_1$, $x_2 \in O_2$ and $O_1 \times O_2 \subseteq O$. As $O_1$ is open in $\Bbb R$, we find $r_1 >0$ such that $$\{y\in \Bbb R: |x_1 - y| < r_1 \} \subseteq O_1\tag{1}$$ and likewise we have $r_2>0$ such that $$\{y\in \Bbb R: |x_2 - y| < r_2 \} \subseteq O_2\tag{2}$$
Now define $r=\min(r_1,r_2) > 0$, and suppose that $d((y_1,y_2), (x_1, x_2)) < r$. Then $$|x_1 - y_1|^2 = (x_1-y_1)^2 \le d((y_1,y_2), (x_1, x_2))^2 < r^2 \le r_1^2$$ 
from which it follows (taking square roots) that
$$|x_1 - y_1| < r_1$$ which implies that $y_1 \in O_1$ by $(1)$. A similar reasoning gives us that $y_2 \in O_2$, and so $(y_1, y_2) \in O$. As $(y_1,y_2) \in B_d((x_1,x_2), r)$ was arbitrary we have shown that 
$$B_d((x_1,x_2), r) \subseteq O$$ and $(x_1, x_2)$ is indeed an interior point for $O$. All in all this shows that $$\mathcal{T}_{\text{prod}} \subseteq \mathcal{T}_d$$
For the other inclusion, we take $O$ $d$-open and show it is product-open, so let $(x_1,x_2) \in O$ again. We need to find $O_1, O_2$ open in $\Bbb R$ such that $(x_1,x_2) \in O_1 \times O_2 \subseteq O$. We know we have $r>0$ such that $$B_d((x_1,x_2), r) \subseteq O\tag{3}$$
Now let $O_1 = \{y \in \Bbb R: |x_1 - y| < \frac{r}{2}\}$ which is open in $\Bbb R$ (it's an open interval/open ball) and similarly is $O_2 = \{y \in \Bbb R: |x_2 - y| < \frac{r}{2}\}$, also open. If $(y_1, y_2) \in O_1 \times O_2$ we know that
$$d((y_1,y_2), (x_1, x_2))^2 = |x_1 - y_1|^2 + |x_2 - y_2|^2 < \frac{r^2}{4} + \frac{r^2}{4} < r^2 $$ which implies (taking roots again)
$$d((y_1,y_2), (x_1, x_2)) < r \text{ and so } (y_1, y_2) \in B_d((x_1,x_2), r) \text{ and by (3) we get } (y_1, y_2) \in O$$
and so $O_1 \times O_2 \subseteq O$ as required.
All in all, this shows $$\mathcal{T}_d \subseteq \mathcal{T}_{\text{prod}}$$
and we have equality of topologies.
