The metric on $\textbf R^{2}$ is defined on pairs of vectors, $x$ and $y$ of two types:
i). $x$ and $y$ are colinear, in which case you subtract them and take the length of the resulting vector.
ii). $x$ and $y$ are not colinear, in which case you add their lengths.
Thus, for example, if $x=(1,1)$ and $y=(3,3)$, $d(x,y)=\sqrt 8$. and if $x=(1,0)$ and $y=(0,1)$, then $d(x,y)=2$.
For the triangle equality, consider cases. First, suppose $x, y$, and $z$ are colinear. Then you verify
$\left \| x-z \right \|\leq \left \| x-y \right \|+\left \| y-z \right \|$, which is true since $\left \| \cdot \right \|$ is the Euclidean metric.
Now suppose $x$ and $z$ are colinear but $y$ and $z$ and therefore $y$ and $x$ are not. Then, you require
$\left \| x-z \right \|\leq d(x,y)+d(y,z)=2\left \| y \right \|+\left \| x \right \|+\left \| z \right \|$.
By assumption, there is a $t\in \textbf R$ such that $x=tz$. So in fact we are checking
$\vert t-1\vert \left \| z \right \|\leq 2\left \| y \right \|+(\vert t\vert +1)\vert \left \| z \right \|$ which means that we need to verify
$(\vert t\vert +1) -\vert t-1\vert )\left \| z \right \|+2\left \| y \right \|\geq 0$, which you can check is true for all $t\in \textbf R$.
Finally, consider the case in which none of the vectors is colinear. Then, we have to see whether
$\left \| x \right \|+\left \| z \right \|\leq \left \| x \right \|+\left \| y \right \|+\left \| y \right \|+\left \| z \right \|$, which is trivial.
Now, for the open sets:
First observe that all they are saying is that if you take a metric space $X$ and restrict your attention to a subset $Y$ of this space, then $Y$ becomes a metric space in its own right, if we use the metric we have on $X$ which will work on $Y$ anyway because $Y$ is a subset of $X$. The fancy way of saying this is that we are giving $Y$ the subspace topology of $X$ induced by $d$. But how must we define the open sets in $Y$? For example, if you take $X=\textbf R$ and $Y=[0,1)\subseteq \textbf R$, you would expect sets like $[0,1/2)$ to be open in $[0,1)$ even though $[0,1/2)$ not open in $\textbf R$. But now observe that
$[0,1/2)=[0,1)\cap (-1,1/2)$. i.e. $[0,1/2)$ IS the intersection of an open set in $\textbf R$ with the subspace we are looking at, $[0,1)$. This gives us a clue as to how we should define the open sets in any subspace:
Let $Y$ be a subset of a metric space $X$. Then, using the same metric on $Y$ that we use on $X$, we declare $G$ open in $Y$ if and only if it is the intersection of some open set $U$ in $X$ with $Y$ itself. That is, if and only if $G=U\cap Y$ for some open set $U$ in $X$.
What you are being asked to do in part two, is to check that this is true. So, let's suppose $G$ is open in the subspace $Y$. We may assume that $G$ is an open ball since by definition of what it means to be open, $G$ is a union of these. So, suppose we have an open ball $B'(y_{0},\epsilon )$, which is just $\left \{ y\in Y:d'(y_{0},y)<\epsilon \right \}$. Then, since $d'$ on $Y$ $=d$ on $X$, we can take $B(y_{0},\epsilon )$ in $X$ now, which is $\left \{ x\in X:d(y_{0},x)<\epsilon \right \}$ and this is a ball in $X$ which contains $B'$. Another way of sayng this is $B\cap Y=B'$ which is what we wanted: an open set in $X$ whose intersection with $Y$ is open in $Y$.
Now suppose $U$ is an open in $X$. We wish to show that the set $G$ defined by $G=U\cap Y$ is open in $Y$. So we need to find an open ball in $Y$ which is contained in $G$.
Now, if the intersection is empty, then it is open trivially. If not, pick an $x_{0}\in G$. Now, $U$ is an open set in $X$ so it contains a ball $B= \left \{ x\in X:d(x_{0},x)<\epsilon \right \}$ as soon as we make $\epsilon $ small enough. Next, by definition of $d'$, we take $B'=\left \{ y\in Y:d(x_{0},y)<\epsilon \right \}$ which is contained in $B$. Because $d=d'$ on $Y$, we have $B'=\left \{ y\in Y:d'(x_{0},y)<\epsilon \right \}$. $B'$ is clearly an open ball in $Y$. Since $B'\subseteq U\cap Y$, we are done.
