First of all: yes, there is already a post about it, but I missread retract as strong deformation retract and wanted to know if this solution is right if we really do assume the stronger assumption of a derformation retract.
Let $T^2$ be the two-dimensional torus and $D$ an embedded disk. We will show by contraposition that $\partial D \subset X$, where $X:=T^2-intD$, is not a deformation retract.
So suppose $\partial D \subset X$ is a strong deformation retract.
Consider the SES (short exact sequence):
$$ 0 \to S_p(D) \stackrel{i}\to S_p(T^2) \stackrel{j}\to S_p(T^2,D)\to0$$
By the exactness axiom (or a theorem about singular homology) we get a LES (long exact sequence) in homology:
$$...\stackrel{\partial ^*}\to H_p(D) \stackrel{i^*}\to H_p(T^2)\stackrel{j^*}\to H_p(T^2,D) \stackrel{\partial ^*}\to H_{p-1}(D) \stackrel{i^*}\to...$$
Since $D$ is contractible we get an isomorphism by exactness in non-zero degrees:
$$H_p(T^2) \cong H_p(T^2,D)\space (\forall p \in \mathbb Z -\{0\})$$
Now we define $U:=intD$ and apply excision on the pair $(T^2,D)$:
$$H_p(T^2,D) \cong H_p(X,\partial D) \space (\forall p \in \mathbb Z -\{0\})$$
But since $\partial D \subset X$ is a strong derformation retract we obtain: $$(X,\partial D) \simeq (\partial D, \partial D) \implies H_2(T^2) \cong H_2(\partial D, \partial D)=0$$
But we know that $H_2(T^2) \cong \mathbb Z \oplus \mathbb Z$ and we have a contradiction.
Correct?
The Solution is correct but more complicated than necessary. One can just show that $X$ and $\partial D$ are not homotopy equivalent by Computing their fundamental group or First homology.
Namely, $\partial D$ is homeomorphic to $S^1$, so $\pi_1=H_1=Z$. But $X=T^2-D^2$ is homotopy equivalent to the wedge of 2 circles, so $\pi_1=Z*Z$ and $H_1=Z\oplus Z$.