First I'll go through my working. Throughout we assume the homology groups of the torus and circle are known.
Let $X=S^1 \times S^1$ be the torus, and $A=S^1 \times \{1\}$. The following is part of a long exact sequence: $$ \dots \to H_n(X) \to H_n(X,A) \to H_{n-1}(A) \to \dots $$ For $n\geq 3$, we have $H_n(X) = H_{n-1}(A) = 0$, so $H_n(X,A) = 0$. Using the 'reduced version' of the sequence, we have $$ \dots \to H_2(A) \to H_2(X) \to H_2(X,A) \xrightarrow{f} H_1(A) \xrightarrow{i_*} H_1(X)\qquad\qquad\qquad\qquad $$ $$ \qquad\qquad\qquad\qquad\qquad \to H_1(X,A)\to \tilde{H}_0(A) \to \tilde{H}_0(X) \to \tilde{H}_0(X,A) \to 0. $$
Step 1. Now, $A, X$ are path connected so $\tilde{H}_0(A) =\tilde{H}_0(X) = 0$. So $\tilde{H}_0(X,A)=0$ by the long exact sequence, which gives us that $H_0(X,A) \cong \mathbb{Z}$.
Step 2. The map $i_*: H_1(A) \to H_1(X)$ takes $[\alpha] \mapsto [i(\alpha)]$ where $i$ is the inclusion. In other words, it is injective, and so $0 = \ker i_* = \text{im}\:f$, and so $f$ is the zero map. This then leaves two short exact sequences, and implies that $H_2(X,A) \cong H_2(X) \cong \mathbb{Z}$.
Step 3. It is an immediate consequence of the other remaining short exact sequence that $H_1(X,A) \cong H_1(X)/\text{im }i_* \cong \mathbb{Z} \oplus \mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}$.
What I'm looking for are mainly comments on if this was done properly, but I also have two specific queries:
1. Have I used the reduced homology 'version' of the long exact sequence correctly? I'm very iffy with reduced homology...
2. My reason for $i_*$ being injective was a little handwavy. Is it always the case that a homomorphism induced by inclusion of spaces $A\hookrightarrow X$ is injective?