Let normed spaces X = (V, $\lVert \cdot \rVert_{\infty}$) and Y = (V, $\lVert \cdot \rVert_{1}$) where $V := \mathcal{C}([a,b]$ (the set of all functions which are continuous on the intervall [a,b]) , $\lVert f \rVert_{\infty} := \max_{x \in[a,b]}(|f(x)|)$ and $\lVert f \rVert_{1} := \int_a^b |f(x)|dx$
I'm asked to comment on the continuity of the identity functions $id : X \rightarrow Y$ as well as $id : Y \rightarrow X$. Taking X and Y as metric spaces by asigning them distances $d_x(f,g) = \max_{x \in[a,b]}(|f(x)-g(x)|)$ and $d_y(f,g) = \int_a^b |f(x)-g(x)|dx$ I can establish the first continuity by saying that by definition:
$\int_a^b |f(x)-g(x)|dx \leq |a-b|\cdot\max_{x \in[a,b]}(|f(x)-g(x)|)$
$d_y(f,g) \leq Cd_x(f,g), \forall f,g\in V^2$ where $C = |a-b|$ which should prove lipschitz continuity.
Im wondering as a more general question if the Lipschitz continuity of a function $f : X \rightarrow Y$ can tell you anything about the Lipschitz continuity of the inverse $f^{-1} : Y \rightarrow X$ by using this as an example. In this case I can prove that the inverse identiy function is not continuous, but only via an unrelated proof that shows it as discontinuous around zero, I can't find a direct link with the lipschitz method I used in the first part of the question.
The function $$ \begin{array}{rccc} f\colon & [0,1] & \longrightarrow & [0,1] \\ & x & \longmapsto & x^2 \end{array} $$ is a $2$-Lipschitz bijection. However, its inverse $$ \begin{array}{rccc} f^{-1}\colon & [0,1] & \longrightarrow & [0,1] \\ & x & \longmapsto & \sqrt{x} \end{array} $$ is not Lipschitz, although it is continuous.
What is true, however, is that a linear map between two normed vector spaces is continuous if and only if it is Lipschitz. So that in your specific case, showing that the identity map (which is linear) is not Lipschitz suffices to say it is not continuous.