Consider the pair $(\Bbb T, A)$ where $A \subset \Bbb T$ is one of the longitudal circles in $\Bbb T$. We thus have the long exact sequence
$$\cdots \to H_{k+1}(\Bbb T) \to H_{k+1}(\Bbb T, A) \stackrel{\partial}{\to} H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \stackrel{\partial}{\to} H_{k-1}(A) \to \cdots$$
$(\Bbb T, A)$ has homotopy extension property, as it's a CW-pair. Thus, we have $H_\bullet(\Bbb T, A) \cong H_\bullet(\Bbb T/A) \cong H_\bullet(S^2 \vee S^1)$ which we know is $\Bbb Z$ if $k = 1, 2$ and is trivial otherwise.
The whole point of choosing this pair is that there is a retract $r : \Bbb T \to A$ which, after applying the $H_\bullet$ functor, gives a left-inverse for the map $H_\bullet(A) \to H_\bullet(\Bbb T)$ induced by the inclusion, which proves injectivity of this map. Thus, the snake maps $\partial$ are all zero
Thus, at each $k$, the long exact sequence reduces to the split (the section being $H_\bullet(r)$) exact sequence
$$0 \to H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \to 0$$
Thus, $H_k(\Bbb T) \cong H_k(A) \oplus H_k(\Bbb T, A)$ which is $\Bbb Z \oplus \Bbb Z$ for $k = 1$, $\Bbb Z$ for $k = 2$ and trivial otherwise.
To add, $H_\bullet(S^2 \vee S^1)$ can be computed in a similar vein, without using Mayer-Vietoris sequence, by taking the CW-pair $(S^2 \vee S^1, S^2)$. There is the retract $S^2 \vee S^1 \to S^2$ by collapsing the circle to the wedged point, which gives a left-inverse for $H_\bullet(S^1) \to H_\bullet(S^2 \vee S^1)$ induced from inclusion as well as a section for the short exact sequence.
Best Answer
Alexander Duality should do the trick.
However, the answer depends on which open set you mean. For example $\mathbb R^2 \setminus I$, where I is the open unit interval has the homology of $S^1$, whereas $\mathbb R^2\setminus \mathbb R \times \{0\}$ has the homology of two points.
In particular, find a bounded open subset $U$ hoeomorphic $\tilde{H_j}(M) \cong \tilde{H}^{n-j-1}(S^n \setminus M)$ for $M$ a closed submanifold (excision will give the desired result.)