I'm reading Weintraub's Fundamentals of Algebraic Topology, in which there is an exercise (3.4.6 in the first edition) that wants us to show that for every path connected subspace $A$ of a path connected space $X$ there is a long exact sequence in reduced homology:
$$\cdots\to \tilde{H}_{n}(A) \to \tilde{H}_{n}(X) \to H_{n}(X,A) \to \tilde{H}_{n-1}(A) \to \cdots. (1)$$
Here $H_*$ is an arbitrary homology theory and the reduced homology groups are defined as
$$\tilde{H}_{n}(Y):=\ker(H_{n}(Y) \to H_{n}(\bullet)).$$
From the chapter that contains the exercise, we know that $\tilde{H}_{n}(Y) \cong H_{n}(Y,y_0)$ for any $y_0 \in Y$ and about the long exact homology sequence for triplets $B\subseteqq A \subseteqq X$ that is
$$\cdots\to H_{n}(A,B) \to H_{n}(X,B) \to H_{n}(X,A) \to H_{n-1}(A,B) \to \cdots. (2)$$
From what I can tell we get the solution immediately by applying the last result to some $B=x_0\in A$. It seems like the author wants us to re-derive this special case of the long exact sequence for triplets using the path-connectedness hypotheses about $A$ and $X$, but I am skeptical that this would work as it is a purely algebraic result.
My question is whether or not if there is a simpler argument to show that (1) exists in this situation, or if I've made a mistake in identifying (1) and (2).
I think the reason for the path-connectedness hypothesis is related to the definition of reduced homology, and not the long exact sequence you are asked to establish: the relative homology groups $H_n(Y,y_0)$ for different $y_0 \in Y$ will be naturally isomorphic if $Y$ is path connected, but if $Y$ has more than one path-component, $H_n(Y,y_0)$ may vary for $y_0$ in different path-components. Thus if you want $\tilde{H}_n(X)$ and $\tilde{H}_n(A)$ to be well-defined, you need to assume that $A$ and $X$ are path-connected.
Update
Apologies -- I misunderstood the OP to be defining $\tilde{H}_*(X) = H_*(A,y_0)$, and Martin's comment is exactly right: taking $\tilde{H}_*(X) = \ker((a_X)_*)$, where $(a_X)_*$ is the map on homology induced by the canonical map $a_X \colon X \to \star$ from $X$ to a one-point set $\star$, the issue of path-connectedness arises when defining $\partial_{\text{red}} \colon H_n(X,A) \to \tilde{H}_{n-1}(A)$: The long exact sequence of a triple gives you the map $\partial_{\text{triple}} \colon H_n(X,A) \to H_{n-1}(A,B)$ hence taking $B = \{y_0\}$ for $y_0\in A$, we obtain $\partial_{\text{red}} \colon H_n(X,A) \to \tilde{H}_{n-1}(A)$ by composing $\partial_{\text{triple}}$ with the isomorphism $r_{y_0}\colon H_{n-1}(A,y_0) \to \tilde{H}_{n-1}(A)$. Since $r_{y_0}$ depends on the path-component of $y_0$ in $A$, $\partial_{\text{red}}$ will thus not be unique unless $A$ is path-connected.
Non-uniqueness: It might be useful to see what happens in a simple example: If $X = X_1\cup X_2\subset\mathbb R^2$ where $$ X_1 = \{(x,\sin(x): 0\leq x\leq 2\pi\}, \quad X_2 \{(x,0): 0\leq x \leq 2\pi\}, $$ and we let $A = X_1\cap X_2= \{(k\pi,0): k\in\{0,1,2\}\}$ and for convenience, let $a_k = (k\pi,0)\in A, (k\in \{0,1,2\})$.
Then $H_n(X)=0$ for $n \geq 2$, $H_1(X)=\mathbb Z^2$ and $H_0(X)=\mathbb Z$, while $H_n(A)=0$ for $n \geq 1$ and $H_0(A) = \mathbb Z^3$ with a basis given by the homology classes of the elements of $A$. The relative homology groups of $(X,A)$ all vanish apart from $H_1(X,A)$, which is isomorphic to $\mathbb Z^2$ (with a basis given by chains $\{[s_0],[s_1]\}$ corresponding to the sine curve between $[0,\pi]$ and $[\pi, 2\pi]$), while the relative homology groups of $(A,\{y_0\}) = (A,\{a_i\})$ (some $i \in \{0,1,2\}$) all vanish apart from $H_0(A,\{y_0\})= H_0(A)/\mathbb Z[a_i]$. The reduced homology of $A$ is $\{\sum_{i=1}^3 c_i[a_i]: \sum_{i=0}^3 c_i =0\}$, and we have isomorphisms $q_i\colon \tilde{H}_0(A) \to H_0(A,\{a_i\})$.
The boundary map $\partial_{\text{red}}\colon H_1(X,A) \to \tilde{H}_0(A)$ should take a chain with boundary in $A$ to a reduced chain in $A$, but the boundary of a chain may not lie in $\ker((a_A)_*)$: we can chose $s_0,s_1$ are as above so that $\partial[s_0] = [a_0]-[a_1]$, $\partial_{\text{pair}} [s_1] = [a_1]-[a_2]$, hence $\partial_{\text{pair}}([s_0]-[s_1]) = -2[a_1] \notin \ker((a_X)_*)$. However if we pick $y_0 = a_2$, the image of $\partial_{\text{triple}}([s_0]-[s_1]) = -2[a_1]+\mathbb Z.[a_2]$, and $q_i^{-1}(-2[a_1]+\mathbb Z.[a_2]) = 2[a_1]-2[a_2]$, giving $\partial_{\text{red}}([s_0]-[s_1]) = 2[a_1]-2[a_2]$. On the other hand, if we take $B=\{a_1\}$ and identify $\tilde{H}_0(A)$ with $H_0(A,\{a_1\})$ then we obtain a boundary map $\partial_{\text{red}}$ for which $\partial_{\text{red}}([s_0]-[s_1]) = 2[a_1] =0$.