The example here is taken from the wikipedia article:
Discontinuous Linear Map
Given the spaces of polynomials $X:=\mathcal{P}([0,1])$ and $Y:=\mathcal{P}([2,3])$.
Their completions being $\hat{X}=\mathcal{C}([0,1])$ and $\hat{Y}=\mathcal{C}([2,3])$.
Consider the operator $T:X\to Y:p\mapsto p$.
Then its closure would be all of $\hat{X}\times\hat{Y}$. Nice!
But is this operator be well-defined: $$p(x)=q(x),x\in[0,1]\implies p(y)=q(y),y\in[2,3]$$ (I guess so as polynomials are analytic and therefore determined by any neighborhood of any point.)
Yes. If you knew $\deg p=d$ then $p$ is uniquely determined by its values at any $d+1$ distinct points. Even if we don't know $\deg p$, it follows that $p$ is uniquely determined by its values on a countable set of points (because that contains $d+1$ points no matter what $d$ is).