The specific answer is very easy. The spectrum of a bounded operator is always a compact subset of $\mathbb C$.Conversely, if we consider for instance bounded operators on a Hilbert space, for any compact $K\subset\mathbb C$ there exists $T\in B(H)$ with $\sigma(T)=K$. In particular, since the Maldelbrot set is compact, it is the spectrum of a certain operator $T\in B(H)$.
The construction is not very exciting, actually. Consider $H=\ell^2(\mathbb N)$ and fix an orthonormal basis $\{e_n\}$. Given $K\subset\mathbb C$ compact, let $\{c_n\}$ be a countable dense subset of $K$. Then define $T\in B(H)$ by
$$
Te_n=c_ne_n
$$
and extend by linearity.
Each $c_n$ is an eigenvalue of $T$ by construction. As the spectrum is closed, the closure of $\{c_n\}$ is in $\sigma(T)$, so $K\subset \sigma(T)$. And if $c\not\in K$, then there exists $\delta>0$ with $|c-c_n|>\delta$ for all $n$. Then, given $x=\sum x_ne_n\in H$,
$$
\|(T-cI)x\|^2=\left\|\sum_n x_n(c_n-c)e_n\right\|^2=\sum_n|x_n|^2\,|c_n-c|^2\geq\delta^2\,\sum_n|x_n|^2=\delta^2\|x\|^2.
$$
Thus $T-cI$ is bounded below; it is easy to show that it is surjective, and so it is invertible. Thus $\sigma(T)=K$.
Note also that $T$ is normal, so we don't need $T$ to be very exotic.
The following argument is actually the same as above, but presented in a different way. Given $K\subset \mathbb C$ compact, we may consider the Banach algebra $C(K)$ (complex-valued continuous functions over $K$). We can see these as linear operators acting (by multiplication) on $L^2(K)$. Then the spectrum of each function is the closure of its range. In particular, the spectrum of the identity function $f(z)=z$ is $K$.
For a densely defined operator $A$ we have, by definition,
$$
D(A^*) = \{y\in H\,|\,\exists z_y\in H\,\forall x\in D(A) : (Ax,y) = (x,z_y)\}.
$$
And if $y\in D(A^*)$, then $A^*y = z_y$. From here you easily see that $R(A)^\perp = N(A^*)$, where $R$ and $N$ denote range and kernel, respectively. Here's how it goes: $y\in N(A^*)$ means $A^*y = 0$, i.e., $y\in D(A^*)$ and $z_y = 0$. This is equivalent to $(Ax,y) = 0$ for all $x\in D(A)$. That is, $y\in R(A)^\perp$.
Now, let $T$ be selfadjoint such that $N(T) = \{0\}$. Then $R(T)^\perp = N(T^*) = N(T) = \{0\}$. Thus, $\overline{R(T)} = H$. Hence, $T^{-1}$ is densely defined (on $R(T)$). We have $T^{-1} : R(T)\to D(T)$. Now,
\begin{align*}
D((T^{-1})^*)
&=\left\{x\in H\,|\,\exists z_x\in H\,\forall y\in R(T) : (T^{-1}y,x) = (y,z_x)\right\}\\
&= \left\{x\in H\,|\,\exists z_x\in H\,\forall u\in D(T) : (T^{-1}Tu,x) = (Tu,z_x)\right\}\\
&=\left\{x\in H\,|\,\exists z_x\in H\,\forall u\in D(T) : (u,x) = (Tu,z_x)\right\}.
\end{align*}
But the latter means exactly that $z_x\in D(T^*) = D(T)$ and $Tz_x = x$. This is equivalent to $x\in R(T) = D(T^{-1})$. Thus, $D((T^{-1})^*) = D(T^{-1})$. Let $x$ be a vector in this set. Then from the first line above you see that $(T^{-1})^*x = z_x$. But we also saw that $Tz_x = x$, i.e., $z_x = T^{-1}x$. This finally implies $(T^{-1})^*x = z_x = T^{-1}x$ and therefore $(T^{-1})^* = T^{-1}$. That is, $T^{-1}$ is selfadjoint.
Best Answer
Consider $X=Y=\ell^2(\Bbb N)$. Let $A(\sum_n x_n e_n)=\sum_n 2^{-n}\,x_ne_n$. $A$ is an injection, however for any $n$ you have $$(A-2^{-n}\Bbb 1)(e_n)=0$$ so $A-2^{-n}\Bbb1$ is never an injection. But for any $\epsilon>0$ you have an $n$ with $A-2^{-n}\Bbb1\in B_\epsilon(A)$. Also $A$ has dense range (since it contains the finitely supported sequences), but you can also verify that $\pi_n (A-2^{-n}\Bbb1)=0$ where $\pi_n$ is the orthogonal projection onto $\mathrm{span}(e_n)$. So the range of $A-2^{-n}\Bbb1$ must lie in the kernel of $\pi_n$, which is a closed subspace of $\ell^2$. So $A-2^{-n}\Bbb1$ also never has dense range.