I already proved that $$ H_{p}(S^{1})=\begin{cases} \mathbb{R} & ,p=0,1\\ 0 & ,\mbox{ otherwise} \end{cases} $$
I now want to prove that $$ H_{p}(S^{2})=\begin{cases} \mathbb{R} & ,p=0,2\\ 0 & ,\mbox{ otherwise} \end{cases} $$
Consider the 2-sphere $S^{2}$ with poles $N=(0,0,1)$ and $S=(0,0,-1)$. Let $U=S^{2}\backslash\{N\}$ and $V=S^{2}\backslash\{S\}$. Notice that $$ U\cap V=S^{2}\backslash\{S,N\}\cong S^{1}\times\mathbb{R}\simeq S^{1} $$
where $\cong$ denotes the homeomorphism relation and $\simeq$ denotes the homotopy equivalence relation.
The Mayer-Vietoris Sequence is
$$ \begin{array}{cccccccc} 0 & \rightarrow & H_{0}(S^{2}) & \rightarrow & H_{0}(U)\oplus H_{0}(V) & \rightarrow & H_{0}(U\cap V) & \rightarrow\\ & \rightarrow & H_{1}(S^{2}) & \rightarrow & H_{1}(U)\oplus H_{1}(V) & \rightarrow & H_{1}(U\cap V) & \rightarrow\\ & \rightarrow & ... \end{array} $$
Now, $H_{0}(S^{2})=\mathbb{R}$ because $S^{2}$ has one connected component. For the same reason, $H_{0}(U)=H_{0}(V)=H_{0}(U\cap V)=\mathbb{R}$.
Also, $U$ and $V$ are homotopically equivalent to $\mathbb{R}^{2}$. Hence,
$$ H_{p}(U)=H_{p}(V)=H_{p}(\mathbb{R}^{2})=\begin{cases} \mathbb{R} & ,p=0\\ 0 & ,\mbox{ otherwise} \end{cases} $$
and, since $U\cap V\simeq S^{1}$, we have $H_{p}(U\cap V)=H_{p}(S^{1})$.
Thus, we can rewrite the MVS as follows:
$$ \begin{array}{cccccccc} 0 & \rightarrow & \mathbb{R} & \rightarrow & \mathbb{R}\oplus\mathbb{R} & \rightarrow & \mathbb{R} & \rightarrow\\ & \rightarrow & H_{1}(S^{2}) & \rightarrow & 0 & \rightarrow & H_{1}(S^{1})=\mathbb{R} & \rightarrow\\ & \rightarrow & H_{2}(S^{2}) & \rightarrow & 0 & \rightarrow & H_{2}(S^{1})=0 & \rightarrow\\ & \rightarrow & 0 & \rightarrow & 0 & \rightarrow & ... \end{array} $$
And now I use a theorem that I read in Tu's book:
Let $0\rightarrow A^{0}\rightarrow A^{1}\rightarrow...\rightarrow A^{m}\rightarrow0$ be an exact sequence of finite-dimensional vector spaces.
Then, $$ \underset{k=0}{\overset{m}{\sum}}(-1)^{k}\dim A^{k}=0 $$
In our case, if we restrict ourselves to the exact sequence $$ \begin{array}{cccccccc} 0 & \rightarrow & \mathbb{R} & \rightarrow & \mathbb{R}\oplus\mathbb{R} & \rightarrow & \mathbb{R} & \rightarrow\\ & \rightarrow & H_{1}(S^{2}) & \rightarrow & 0 & \rightarrow & H_{1}(S^{1})=\mathbb{R} & \rightarrow\\ & \rightarrow & H_{2}(S^{2}) & \rightarrow & 0 \end{array} $$ we get
$$ 1-2+1-x_{1}+0-1+x_{2}=0 $$
with $x_{1}=\dim H_{1}(S^{2})$ and $x_{2}=\dim H_{2}(S^{2})$.
So $$ x_{2}-x_{1}=1 $$
One option is obviously $x_{2}=1$ and $x_{1}=0$. But how can I show that the other options are incorrect? Also, just because $\dim H_{2}(S^{2})$ does that mean that $H_{2}(S^{2})=\mathbb{R}$? (I think I saw this argument being made before). Why?
Part of your long exact sequence is $$0\to H^1(S^1)=\Bbb R\to H^2(S^2)\to0.$$ This already shows that $H^1(S^1)$ and $H^2(S^2)$ are isomorphic, so $x_2=1$.