The Toeplitz theorem says : If a closed operator is defined everywhere, then it is continuous.
So if a non continuous operator is defined everywhere, it is not closed. But why is it not closed? What condition fails for it to be closed? I thought if a (non continuous) operator $A$ is symmetric and defined on a dense domain in $H$, we can always close it (unambiguously) :
For $f$ where $Af$ is not defined : if $f_n \rightarrow f$ and $\lim Af_n$ exists (i.e. $A f_n \rightarrow f^*$), we define $\bar A f = f^*$ (It will be well defined because if $f_n\rightarrow f,g_n \rightarrow f,Af_n \rightarrow f^*, Ag_n \rightarrow g^*$ then $f^*=g^*$). For all other $f$, we define $\bar Af = Af$.
So $\bar A$ will then be a closed operator defined everywhere, and if it's non continuous that would contradict the theorem. Does that mean that $\bar A$ will be continuous?
When you close a densely defined symmetric operator $A$, you get a closed operator $\bar{A}$, but $\bar{A}$ will (usually) still not be defined on all of $H$. There will simply exist some $f\in H$ such that there exists no sequence $(f_n)$ in the domain of $A$ converging to $f$ such that $\lim Af_n$ exists, so your definition of $\bar{A}$ fails for $f$. Indeed, Toeplitz's theorem says that if $A$ is unbounded, then some such $f$ must exist.