This question is somewhat similar to Form functions that are continuous at one point in L^\infty a Banach space. where the continuity at zero was added to $L^\infty(\Omega)$. This space was indeed complete.
I want to consider $$X=\{u \in H^1(\Omega) : \text{$u$ is continuous at $0$}\}$$ with $0 \in \Omega \subset \mathbb{R}^2$ (i.e., $H^1(\Omega)$ is not embedded in the space of continuous functions) and equip it with the norm $$\|u\|_X^2 = \|u\|_{H^1(\Omega)}^2 + |u(0)|^2$$ This is indeed a norm and we can also derive a scalar product from it. I want to know whether this space is complete. So, let us take a Cauchy sequence $u_n$ in $X$. Then, $u_n$ is a Cauchy sequence in $H^1(\Omega)$ and converges to $u$ in $H^1(\Omega)$. Now, I have to show that $u_n(0)$ converges and it indeed converges to $u(0)$. Is this true?
No, this is not true. Using the usual arguments, one can construct a sequence $u_n \in H^1(\Omega) \cap C(\Omega)$ with $u_n \to 0$ in $H^1(\Omega)$, but $u_n(0) = 1$ for all $n \in \mathbb N$.
Let $\varphi \in H^1(\Omega)$ be given, such that $\varphi$ has a singularity at $0$, but is otherwise continuous. Then, we can use $$ u_n := \min( \varphi/n, 1). $$ It is straightforward to verify, that this sequence has the desired properties.