Identity map between metric spaces continuous or not. How to counter this problem? Is it enough to show pre image of unit ball in some metric is open under another to show the continuity? I am not at all getting the path to proceed. And how to contradict when it's not continuous?

 A: We know that a continuous map between metric spaces is one which preserves limits: $f$ is continuous if and only if $\lim\limits_{n\to\infty} f(x_n)=f(\lim\limits_{n\to\infty} x_n)$ for each convergent sequence $x_n$. To prove that a map is not continuous, we can try to find a sequence that is convergent in one metric but not the other.
For example, I don't think that b) is true. Consider the sequence
$$f_n(x)=x^n.$$
Then $\lim\limits_{n\to\infty} f_n=0$ in the metric $d_2$, but not in $d_1$: it even fails to be Cauchy in $d_1$.
Of course, in general it is fine to show that the preimage of any open ball is open. There is some general theory that tells you that if the identity map is bounded - that is, if for every $x\in X_i$, $d_j(x,0)\le M\cdot d_i(x,0)$ -- then the identity map from $X_i$ to $X_j$ is continuous. This uses the fact that all these metrics come from norms, though, and takes some work to establish.
A: a. Let $\varepsilon>0$ be given. Since $\sup_{x\in [0,1]} | \, f(x)-g(x)| \geq \int_0^1 |\, f(x)-g(x)|dx$, we may choose $\delta=\varepsilon$ so that $\int_0^1 |\, f(x)-g(x)|dx<\varepsilon$ whenever $\sup_{x\in [0,1]} |\, f(x)-g(x)|<\delta$. Therefore $id:X_1 \longrightarrow X_2$ is continuous.
b. Notice $\int_0^1 |x^n|dx \to 0$ as $n \to \infty$ BUT $\sup_{x\in [0,1]} |x^n|=1$ for every $n \in \mathbb N$. Therefore $id:X_2 \longrightarrow X_1$ is NOT continuous at the constant function $g \equiv 0$ because it is not sequentially continuous there. There are other counterexamples but this is the preferred (by at least 2 people in this thread).
c. I leave to you. The notion of sequential continuity seems good for demonstrating a counterexample to such statements, while the standard notion of continuity seems good for proving such statements true which is not exactly strict advice.

Technically for b. I used the fact that the functions $f_n(x):=x^n$ from the sequence of functions $\{f_n\}_{n=1}^\infty$ converge to the constant function $g \equiv 0$ in $(X_2, d_2)$ but the same is not true in $(X_1, d_1)$. And so $id: X_2 \longrightarrow X_1$ is not sequentially continuous at $g$ as opposed to identifying the pointwise limit $f(x):=\lim_{n\to \infty} f_n(x)$ which is not in the set $\mathcal C[0,1]$ as $f$ is not continuous at $x=1$. 
If this seems like an issue, then showing that the chosen sequence of functions is not uniformly Cauchy on $[0,1]$ (Cauchy in $(X_1, d_1)$) avoids this. For this argument we make the following two observations, $|x^n-x^m|=|x^n||1-x^{m-n}|$ and the function $F(x):=\left(1-\frac1x\right)^x$ is nondecreasing on $[1, \infty)$ with $\lim_{x \to \infty} F(x)=\frac1e$. So with $\varepsilon=\frac1{4}(1-e^{-1})$ and for any $n\geq 2$, we may select $m=2n$ and have
$$\sup_{x\in[0,1]} |x^n-x^m| \geq \left(1-\frac1n\right)^n\left(1-\left(1-\frac1n\right)^n\right)\geq \varepsilon$$
which shows that the sequence of functions $\{f_n\}_{n=1}^\infty$ is not Cauchy in $(X_1, d_1)$ by definition. 
