Show that the metric $d$ induces the euclidean topology on $\mathbb R^2$ 
Consider $\mathbb R^2$ with a metric given by $d((a_1,a_2),(b_1,b_2))=|a_1 - b_1|+ |a_2 -b_2|$.

I need to show that the metric $d$ induces the euclidean topology on $\mathbb R^2$, but I am not sure how I can show this. Can anyone please help me show this?

Two metrics $d,d'$ on a set $X$ induce the same topology iff for all $x\in X$ and  all  $r>0, $ we have 
(1). there exists $s>0 $ such that $B_d(x,s)\subset B_{d'}(x,r)$,.... and
(2). there exists $s'>0$ such that $B_{d'}(x,s')\subset B_d(x,r).$
On $R^2, $ with $d((a,b),(c,d))=((a-c)^2+(b-d)^2)^{1/2} $ and $d'((a,b),(c,d))=|a-c|+|b-d|, $  we may put $s=r/\sqrt 2$ in (1) and $s'=r/\sqrt 2$ in (2). It helps here to draw a diagram.
In general , when $B,B'$ are bases respectively for topologies $T, T'$ on a set $X, $ suppose that  whenever $x\in b\in B, $ there exists $b'\in B'$ with $x\in b'\subset b.$ Then for any $b\in B$ and any $x\in b, $ $$\text {let }\quad  C(x,b)=\{b'\in B': x\in b'\subset b\}.$$ $$\text {And let }\quad D(b)= \cup_{x\in b}\{C(x,b):x\in b\}.$$ We have $D(b)\subset B'\subset T'$ so $b=\cup D\in T'.$ Since this holds for all $b\in B,$ we have $B\subset T' , $ so $T\subset T'.$
If the preceding paragraph also holds with $B,B'$ interchanged and $T,T'$ interchanged, we also have $T'\subset T.$ Of course $T\subset T'\subset T\implies T=T'.$
In the case where the  topologies $T,T'$ are generated by metrics $d,d'$ we may take $B$ and $B'$ to be the families of all open $d ,  d'$ balls, respectively, which brings us to my first sentence.
