Let $X=C[0,1]$ and define $I:(X,\|\cdot\|_{\infty})\longrightarrow (X,\|\cdot\|_1 )$. Need to show that $I$ is not open.
I was thinking to find an open set $A$ in $(X,\|\cdot\|_{\infty})$ whose image $I(A)=A$ is not open in $(X,\|\cdot\|_1)$.
But does that make sense to find a set that is open with respect to a norm but not open with respect to another norm, using the same Banach space $X$. Or $I$ may look for another method to prove this.
Yes, your approach makes sense. There will indeed be sets that are open in one norm, but which are not open in the other.
In fact, $A = B_\infty(0; 1)$ should be a counterexample, if a counterexample exists. If $B_\infty(0; 1)$ is open with respect to $\|\cdot\|_1$, then the same is true for every $\infty$-ball. Since $\infty$-open sets are unions of $\infty$-balls, then this would make every $\infty$-open set also $1$-open, and $I$ would be an open map after all.
You can argue even further that $0 \in B_\infty(0; 1)$ cannot be contained in the $1$-interior. So, what would help is to show that $0$ is the $1$-limit of a sequence of functions in $C[0, 1] \setminus B_\infty(0; 1)$.
How do we construct such functions? It's not hard. The $1$-norm doesn't care about how large the function values get, so long as the size of the integral is managed. Functions with a sharp spike in the graph can have small norms, provided the spike is very thin. We need to construct functions with thin enough spikes so that the integral approaches $0$, while the peaks of the spikes give us function values larger than $1$.
With this in mind, let us define, for $n \ge 1$, $$f_n(x) = \begin{cases} nx & \text{if } 0 \le x \le \frac{1}{n} \\ 2 - nx & \text{if } \frac{1}{n} < x \le \frac{2}{n} \\ 0 & \text{otherwise.}\end{cases}$$
Verify that $f_n$ is continuous. Note that $f_n(1/n) = 1$, which means that $\|f_n\|_\infty \ge 1$. But, \begin{align*} \|f_n\|_1 &= \int_0^1 |f_n(x)| \, \mathrm{d}x \\ &= \int_0^{\frac{1}{n}} nx \, \mathrm{d}x + \int_{\frac{1}{n}}^{\frac{2}{n}} 2 - nx \, \mathrm{d}x + \int_{\frac{2}{n}}^1 0 \, \mathrm{d}x \\ &= \left[\frac{n}{2}x^2\right]_0^{\frac{1}{n}} + \left[2x - \frac{n}{2}x^2\right]_{\frac{1}{n}}^{\frac{2}{n}} \\ &= \frac{1}{2n} + \frac{2}{n} - \frac{3}{2n} = \frac{1}{n}, \end{align*} which tends to $0$. That is, $f_n \to 0$ in the $1$-norm, completing the proof.
The other way you could view this is by looking at the inverse map. Since $I$ is bijective, $I^{-1}$ will be continuous if and only if $I$ is open. Since $I^{-1}$ is linear, it suffices to show it's not bounded. So, we should be able to find a sequence of points in $B_1(0; 1)$ whose $\infty$-norm is unbounded. You can give that one some thought yourself.