Homology of the closed topologist sine curve The closed topologist sine curve, $X$, is the subspace of $R^2$ consisting of all the points $(x,\sin(1/x))$ for $x \in (0,1]$, and all points $(0,y)$ for $y \in [-1,1]$ and an arc from $(0,-1)$ to $(1,\sin(1))$ Compute the singular homology of $X$, using a suitable Mayer-Vietoris sequence.
So to utilize the Mayer-Vietoris sequence I need an excisive couple, one condition that will gaurentee that a pair is an excisive pair is if the union of their interiors cover the space. 
$X_1$ = $(x,\sin(1/x))$ for $x \in (0,1]$ and the arc from $(0,-1)$ to $(1,\sin(1))$ 
$X_2$ = all points $(0,y)$ for $y \in [-1,1]$ and the arc from $(0,-1)$ to $(1,\sin(1))$
I don't know i'm just taking a shot in the dark really I guess. Can anyone offer some insight for me? Thanks!!

You do not consider the closed topologist's sine curve, but the Warsaw circle. See To show that Warsaw circle is simply connected.
Let $S = \{(x,\sin\frac{1}{x}):0<x \le 1 \}$, $L = \{0\} \times [-1,1]$ and $T = L  \cup S$. This is the closed topologist's sine curve. Your space $X$ is obtained by joining $(0,-1), (1,\sin(1)) \in T$ by an arc $A$ such that $A$ does not meet other points of $T$.
Your subspaces $X_1, X_2$ do not form an excisive couple. The points of $L \subset X_2$ are no interior points of $X_1$ and $X_2$.
Choose a homeomorphism $h : [0,1] \to A$. Let $X_1 = h((0,1))$ and $X_2 = X \setminus \{h(1/2)\}$. Then you have an excisive couple. $X_1$ is contractible, thus
$$H_n(X_1) = \begin{cases} 0 & n > 0 \\ \mathbb Z & n = 0\end{cases}$$
The space $X_2$ contains $T$ as a strong deformation retract and $T$ has two path components ($L$ and $S$) which are both contractible. Hence
$$H_n(X_2) \approx H_n(T) \approx H_n(L) \oplus H_n(S) = \begin{cases} 0 & n > 0 \\ \mathbb Z \oplus \mathbb Z & n = 0\end{cases}$$
Moreover, $X_1 \cap X_2$ is the disjoint union of two copies $I_1, I_2$ of an open interval. Thus
$$H_n(X_1 \cap X_2) \approx H_n(I_1) \oplus H_n(I_2) = \begin{cases} 0 & n > 0 \\ \mathbb Z \oplus \mathbb Z & n = 0\end{cases}$$
Finally, since $X$ is path-connected
$$H_0(X) = \mathbb  Z .$$
By Mayer-Vietoris we get for $n > 0$ an exact sequence
$$0 \to H_n(X) \to H_{n-1}(X_1 \cap X_2) \to H_{n-1}(X_1) \oplus H_{n-1}(X_2) \to H_{n-1}(X) \to \ldots$$
For $n > 1$ we know that $H_{n-1}(X_1 \cap X_2) = 0$, thus $H_n(X) = 0$.
Let us come to $n = 1$. We get the exact sequence
$$0 \to H_1(X) \to H_0(X_1 \cap X_2) \to H_0(X_1) \oplus H_0(X_2) \to H_0(X) \to 0$$
If $K$ denotes the kernel of $H_0(X_1) \oplus H_0(X_2) \to H_0(X)$, we get a short exact sequence
$$0 \to K \to H_0(X_1) \oplus H_0(X_2) \to H_0(X) \to 0$$
Since $H_0(X) = \mathbb Z$, this sequence splits and we get
$$H_0(X_1) \oplus H_0(X_2) \approx K \oplus H_0(X) .$$
We already know $H_0(X_2) = \mathbb Z \oplus \mathbb Z$ and $H_0(X_1) = H_0(X) = \mathbb Z$, thus
$$\mathbb Z \oplus \mathbb Z \oplus \mathbb Z \approx K \oplus \mathbb Z .$$
This shows that $K \approx \mathbb Z \oplus \mathbb Z$. But $K$ agrees with the image of $H_0(X_1 \cap X_2) = \mathbb Z \oplus \mathbb Z \to H_0(X_1) \oplus H_0(X_2)$. This is only possible if this map is injective. Thus we must have $H_1(X) = 0$.
