$I: (X,d_0) \to (X,d_1) : x \mapsto x$. Is I continuous? $X=[0,1]$, $d_0 = |x| + |y|$ $ x \neq y $ and $d_0(x,x) = 0 $
$d_1 = |x-1|+|y-1|$ $x\neq y$ $d_1(x,x) = 0$
$I: (X,d_0) \to (X,d_1) : x \mapsto x$. Is I continuous?
Definition: $T:(X,d_x) \to (Y,d_y)$ is continuous at $a\in A$ if $ x \to^{d_x} a \implies  Tx \to^{d_y} Ta$
In my notes I wrote I is continuous but according to this definition now it seems to me that it is not since
$1/n \to^{d_0} 0 $ but with metric $d_1$ it does not approach to 0. Thus it is not continuous. 
 A: Your example is correct:
$\displaystyle\frac1n \xrightarrow{d_0}0$ because $$d_0\left(\frac1n, 0\right) = \left|\frac1n\right| + |0| = \frac1n \xrightarrow{n\to\infty} 0$$
$\displaystyle\frac1n \not\xrightarrow{d_1}0$ because $$d_1\left(\frac1n, 0\right) = \left|\frac1n - 1\right|  + |0-1| = 2-\frac1n\xrightarrow{n\to\infty} 2$$
Therefore $I$ is not continuous.

Notice that continuity of $I$ actually means that for every $\varepsilon > 0$ and an open ball $B_{d_1}(x_0, \varepsilon)$ there exists $\delta > 0$ such that $$B_{d_0}(x_0, \delta) = I(B_{d_0}(x_0, \delta)) \subseteq B_{d_1}(x_0, \varepsilon)$$
This is equivalent to the statement that the topology generated by $d_0$ is finer that the topology generated by $d_1$.
You can directly verify that this is not true.
Let's determine $B_{d_1}\left(0, \frac12\right)$.
$$x \in B_{d_1}\left(0, \frac12\right), x \ne 0 \iff \frac12 > d_1(0,x) = |x-1| + |0-1| = 2-x \iff x> \frac32 $$
Therefore, $B_{d_1}\left(0, \frac12\right)  =\{0\}$. But, there does not exist $r > 0$ such that $B_{d_0}\left(0, r\right) \subseteq B_{d_1}\left(0, \frac12\right) = \{0\}$.
$$x \in B_{d_0}\left(0, r\right), x\ne 0 \iff r > d_0(0, x) = |x|+ |0| = x  \iff x \in [0, r\rangle$$
So $B_{d_0}\left(0, r\right) = [0, r\rangle \not\subseteq \{0\} = B_{d_1}\left(0, \frac12\right)$ for all $r > 0$.
