$\mathscr{B}=\{(a,b)\cap\Bbb Q:a,b\in\Bbb R\text{ and }a<b\}$ is indeed a base for the topology that $\Bbb Q$ inherits from the usual topology on $\Bbb R$. After that, though, you’ve gone badly astray. If $p$ and $q$ are distinct rational numbers with $p<q$, then the union of $(\leftarrow,p)\cap\Bbb Q$ and $(q,\to)\cap\Bbb Q$ is definitely not all of $\Bbb Q$: it misses every rational number in the interval $[p,q]$.
It’s true that $\Bbb Q$ is a countable set, but no matter how you index it as $\{q_n:n\in\Bbb N\}$, it won’t be true that $q_n$ and $q_{n+1}$ are adjacent: no two rational numbers are adjacent. If $p$ and $q$ are distinct rational numbers, then $\frac12(p+q)$ lies strictly between them and is also rational.
In order to get a simple disconnection of $\Bbb Q$, pick an irrational number $\alpha$; $\sqrt2$ will do nicely. Then let $U=(\leftarrow,\alpha)\cap\Bbb Q$ and $V=(\alpha,\to)\cap\Bbb Q$; $(\leftarrow,\alpha)\cup(\alpha,\to)=\Bbb R\setminus\{\alpha\}$, and $\alpha\notin\Bbb Q$, so $U\cup V=\Bbb Q$, and certainly $U\cap V=\varnothing$. Finally, if $p\in U$, then $p\in(p-1,\alpha)\subseteq U$, where $(p-1,\alpha)\in\mathscr{B}$, so $U$ is open in $\Bbb Q$. Similarly, if $p\in V$, $(\alpha,p+1)$ is a member of $\mathscr{B}$ containing $p$ and contained in $V$, so $V$ is also open in $\Bbb Q$.
You can do even more: for any two rational numbers $p$ and $q$, there is an irrational number $\alpha$ strictly between them, so you can find a separation of $\Bbb Q$ that puts $p$ and $q$ into opposite sides of the separation. Thus, $\Bbb Q$ is totally disconnected.
To show that $\tau_d\supseteq\tau_1$ it’s not enough to show that each $d_1$-open ball contains a $d$-open ball: you must show that it contains a $d$-open ball with the same centre. You seem to have tried to do this, but it needs to be said as well, and your argument isn’t quite right, though you have found a $d$-open ball that works. All you have to show is that $B_x^d(\delta)\subseteq B_x^{d_1}(\delta)$. To do this, suppose that $y\in B_x^d(\delta)$; then $d_1(x,y)=d(x,y)-d_2(x,y)\le d(x,y)<\delta$, since $d_2(x,y)\ge 0$, so $y\in B_x^{d_1}(\delta)$. As you say, a similar argument yields the conclusion that $\tau_d\supseteq\tau_2$.
For (b) suppose that $U\in\tau_d$; we need to show that $U\in\tau_1$. Let $x\in U$; by hypothesis there is an $\epsilon>0$ such that $B_x^d(\epsilon)\subseteq U$, and we want a $\delta>0$ such that $B_x^{d_1}(\delta)\subseteq U$. We know that $B_x^{d_2}(\epsilon/2)\in\tau_1$, so there is a $\delta_1>0$ such that $B_x^{d_1}(\delta_1)\subseteq B_x^{d_2}(\epsilon/2)$. Let $\delta=\min\{\delta_1,\epsilon/2\}$; if $y\in B_x^{d_1}(\delta)$, what can you say about $d(x,y)$?
Best Answer
Let $(\Bbb R,\tau_e)$, the real numbers with the standard euclidean topology. Since $\Bbb Q$ is dense in $\Bbb R$, it is easy to verify that a base for the euclidean topology is $\mathcal B=$ {$B(y,R)|y\in\Bbb Q,R\in\Bbb Q_{>0}$}.A very trivial fact to verify is that if $\mathcal B$is a base for $X$ and $Y\subset X$, $\mathcal B_y$={$\mathcal B\cap Y|B\in\mathcal B$}is a base for the subspace topology. Another fact is that if $d:X\times X\to[0,+\infty)$ induces a topology on $X$ and given $Y\subseteq X$, if $d_y:Y\times Y\to[0,+\infty)$ is the restriction of $d$ to $Y$, a base for the subspace topology of $Y$ is $\mathcal B=$ {$B(y,R)\cap Y |y\in\Bbb Q,R\in\Bbb Q_{>0}$}. So the subspace of topology inherited from the real numbers is the same as the topology induced by the restriction. So with your notation $\tau_1=\tau_2$. $\square$