The Closedness in the hypothesis of Tietze extension theorem is necessary

Question

Consider $B=\{ (x,y) \in \Bbb{R}^2:x^2+y^2 \leq 1\}$ and $D=\{ (x,y) \in \Bbb{R}^2:x^2+y^2<1\}$.

Prove or disprove:

1) Given a continuous function $f:B \rightarrow \Bbb{R}$, there exist a continuous function $g:\Bbb{R}^2 \rightarrow \Bbb{R}$ such that $g=f$ on $B$.

2) Given a continuous function $s:D \rightarrow \Bbb{R}$, there exist a continuous function $t:\Bbb{R}^2 \rightarrow \Bbb{R}$ such that $t=s$ on $D$.

I know 1) follows from Tietze extension theorem, since $B$ is closed and $\Bbb{R}$ is a normal space.

What about 2? Any hint must be appreciated!

Answer 1

As a simple example where 2. fails: let $s: D \to \mathbb{R}$ be given by $s(x) = \frac{1}{x^2 + y^2 - 1}$. You cannot even continuously extend it to a single point of $B \setminus D =S^1$.