Assume $T$ has empty spectrum. Then $T$ is invertible, $T^{-1}$ is a bounded selfadjoint operator and, for $\lambda \ne 0$,
$$
(T^{-1}-\lambda I) =(I-\lambda T)T^{-1}=\lambda(\frac{1}{\lambda}I-T)T^{-1}
$$
has bounded inverse
$$
\frac{1}{\lambda}T\left(\frac{1}{\lambda}I-T\right)^{-1}
$$
So $\sigma(T^{-1})=\{0\}$ because only $\lambda=0$ can be in the spectrum, and it cannot be empty. But that implies $T^{-1}=0$, which is a contradiction.
Consider the space $H = L^2([0,1],\mu)$, where $\mu$ is the Lebesgue measure. Define $T$ to be the multiplication with the identity function, i.e.
$$(Tf)(x) = x\cdot f(x).$$
Since the identity function is bounded, $T$ is bounded ($\lVert T\rVert \leqslant 1$), and since it is real-valued, $T$ is self-adjoint.
Clearly $T$ has no eigenvalues, since
$$(T - \lambda I)f = 0 \iff (x-\lambda)\cdot f(x) = 0 \quad \text{a.e.},$$
and since $x-\lambda$ is nonzero for all but at most one $x\in [0,1]$, it follows that $f(x) = 0$ for almost all $x$.
We have $\sigma(T) = [0,1]$, as is easily seen - for $\lambda \in \mathbb{C}\setminus [0,1]$, the function $x \mapsto \frac{1}{x-\lambda}$ is bounded on $[0,1]$.
The reason that $T$ has no eigenvalues is that for all $\lambda\in \mathbb{C}$ the set $\operatorname{id}_{[0,1]}^{-1}(\lambda)$ has measure $0$. If $m \colon [0,1] \to \mathbb{R}$ is a bounded measurable function such that $\mu(m^{-1}(\lambda)) > 0$ for some $\lambda$, then $\lambda$ is an eigenvalue of $T_m \colon f \mapsto m\cdot f$, any function $f\in H$ that vanishes outside $m^{-1}(\lambda)$ is an eigenfunction of $T_m$ then. Conversely, if $T_m$ has an eigenvalue $\lambda$ and $f \neq 0$ is an eigenfunction to that eigenvalue, then, since $(m(x) - \lambda)f(x) = 0$ almost everywhere, it follows that $f$ vanishes almost everywhere outside $m^{-1}(\lambda)$, and since $f\neq 0$, it further follows that $\mu(m^{-1}(\lambda)) > 0$.
Best Answer
We want to prove that if $X$ is a Banach space and $T^*\in B(X^*)$ is invertible, then $T$ is invertible in $B(X)$.
We go through a few steps.
Note that $\operatorname{ran} T$ is closed. indeed, let $W$ be the inverse of $T^*$, and let $\{Tx_n\}$ be a Cauchy sequence. Then \begin{align} \|x_n-x_m\| &=\sup\{|f(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|T^*Wf\,(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\sup\{|(Wf)\,(Tx_n-Tx_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &\leq\|Tx_n-Tx_m\|\,\sup\{\|Wf\|:\ f\in X^*,\ \|f\|=1\}\\ \ \\ &=\|W\|\,\|Tx_n-Tx_m\|. \end{align} So $\{x_n\}$ is Cauchy; there exists $x\in X$ with $x=\lim x_n$. As $T$ is bounded, $Tx=\lim Tx_n$, and $\operatorname{ran} T$ is closed.
$T$ is injective. Indeed, if $Tx=0$, then for any $f\in X^*$ we have $f=T^*g$ (since $T^*$ is surjective). Then $$f(x)=T^*g(x)=g(Tx)=g(0)=0.$$ Thus $f(x)=0$ for all $f\in X^*$, and so $x=0$.
$T$ is surjective. Indeed, if $y\in X\setminus \operatorname{ran} T$, using Hahn-Banach (and the fact that $\operatorname{ran} T$ is closed) there exists $g\in X^*$ with $g(y)=1$, $g(Tx)=0$ for all $x$. But then $0=g(Tx)=T^*g(x)$ for all $x$, and so $T^*g=0$. As $T^*$ is injective, $g=0$; this is a contradiction. So $X=\operatorname{ran} T$, and $T$ is surjetive.
Finally, since $T$ is bijective and bounded, by the Inverse Mapping Theorem it is invertible.