Proving that a function is not a contraction

Take $f \equiv 1$ and $g\equiv 0$. Then, $$d_\infty(Tf,Tg) = 1 = d_\infty(f,g).$$

Let $E:=\mathcal{C}^0([a,b],d_\infty)$ for any $-\infty<a<b<+\infty$. Then $(E,d_\infty)$ is a Banach space, so assuming by contradiction that $T$ is a contraction, there exists an unique fixed point $f$ for $T$ by Banach's fixed point theorem.

Take $\alpha>0$ and let $g_\alpha\in E$ be defined by $g_\alpha(x):=\mathrm{e}^{\alpha x}$. Then, as $Tf=f$, we get: $$d_\infty(f-Tg_\alpha)=d_\infty(Tf-Tg_\alpha)<d_\infty(f-g_\alpha).$$ By continuity in $\alpha>0$ of the LHS and the RHS above, and by boundedness of $f$ over $[0,1]$, we have $$d_\infty(f-Tg_\alpha)=\sup_{x\in[0,1]}\left|f(x)-\frac{\mathrm{e}^{\alpha x}}{\alpha}\right|\underset{\alpha\to0^+}{\longrightarrow}+\infty$$ while $$d_\infty(f-g_\alpha)=\sup_{x\in[0,1]}\left|f(x)-\mathrm{e}^{\alpha x}\right|\underset{\alpha\to0^+}{\longrightarrow}\sup_{x\in[0,1]}\left|f(x)-1\right|<+\infty,$$ a contradiction.

The conclusion is therefore that $T$ can not be a contraction on $E$.

EDIT: In fact, $T$ has an unique fixed point which is the trivial function $f=0_E$, see there: Why $f(x)=e^x$ is a fixed point of $T$? The above argument can be adapted to situations where the fixed point, if it exists, is not trivial.