Let $d_{(1)}(x,y)= \sum_{i=0}^n |x_i-y_i|$ and $d_{(2)}(x,y)=\sqrt{\sum_{i=1}^n|x_i-y_i|^2}$, and $d_{(\infty)}(x,y) = sup{|x_i-y_i|}$.
Prove that for any $x,y \in \mathbb R^n$ we have $d_{(2)}(x,y) = d_{(1)}(x,y) $ .
Proof: Let $x,y \in \mathbb R^n$. Then $ (d_{(2)}(x,y))^2=(\sqrt{\sum_{i=1}^n|x_i-y_i|^2})^2= \sum_{i=1}^n|x_i-y_i|^2$.Then $(d_{(1)}(x,y))^2= \sum_{i=1}^n|x_i-y_i|^2$. Therefore we have that $(d_{(2)}(x,y))^2 \le (d_{(1)}(x,y))^2$.
Is this the correct way to prove this statement or is there another way?
Also, I'm really confused as how to start with showing that $d_{(\infty)}(x,y) \le d_{(2)}(x,y).$
Any advice to help give me a jump start with the problem would really be helpful.
The statement is wrong. Indeed:
\begin{align*} d_1(x,y)^2&=\left(\sum_{i=1}^n|x_i-y_i|\right)^2\\ &=\left(\sum_{i=1}^n|x_i-y_i|^2\right)+2\sum_{1\le i<j\le n}|x_i-y_i|\times|x_j-y_j| \end{align*}
The second term is not necessarily 0. If you take for example $x=(1,\dots,1)$ and $y=(2,\dots,2)$, you'll see that $d_1(x,y)=n$ while $d_2(x,y)=\sqrt{n}\ne n$ for $n\ge 2$.
However, notice that $\sum_{1\le i<j\le n}|x_i-y_i|\times|x_j-y_j|\ge0$. This will give you the inequality $d_1(x,y)\ge d_2(x,y)$.
For the second inequality, as cdipaolo said in the comments, take out all the terms of $d_2(x,y)^2$ except the largest one.