From the first condition, you can deduce that $T$ is bounded, using uniform boundedness principle:
$$
\sup_{||x||=1}|f(T(x))|<\infty,\forall f\in E^*
$$
$$
\implies ||f(T)||<\infty,\forall f\in E^*
$$
$$
\implies ||T||<\infty
$$
first step by definition of $||\cdot||$, second by UBP (there's probably a quicker way to show this, but it's been a while since FA).
The second hypothesis will allow you to conclude the operator is bounded below, and in turn, that the range is closed:
if $Tx_n \to y$
$$
|T(x_m) - T(x_n)| > C|x_m - x_n|
$$
so $x_n \to x$ by completeness. Since $Tx_n \to Tx$, we have $Tx = y$.
The third hypothesis prevents the range from being a (non-trivial) closed subspace (if it were, you could find a non-zero $f$ such that $f(y) = 0$ for all $y$ in the subspace, by Hahn-Banach).