Proof:

The hypothesis implies that the pair ${\cal U} = \{$int$\;A,\; X - \overline{U}\}$ is an open cover of $X$, where $\overline{U}$ denotes the closure of $U$. Likewise ${\cal V} = \{$int$\;A,\; A - \overline{U}\}$ is an open cover of $A$ and $S^{\;{\cal V}}(A)$ is a subcomplex of $S^{\;{\cal U}}(X)$. By theorem (29.6) the short exact sequence of complexes

$$\begin{CD} 0@> >> S^{\;{\cal V}}(A) @> >> S^{\;{\cal U}}(X) @> >> S^{\;{\cal U}}(X)/S^{\;{\cal V}}(A) @> >> 0 \end{CD}$$

gives rise to a long exact sequence in homology:

$$\begin{CD} @> >> H^{\;{\cal V}}_n(A) @> >> H^{\;{\cal U}}_n(X) @>... ...{\cal U}}(X)/S^{\;{\cal V}}(A)) @> >> H^{\;{\cal V}}_{n-1}(A)@> >> \end{CD}$$

On the other hand there is an obvious map of complexes induced by the inclusion maps namely

$$\begin{CD} j\;:\;S^{\;{\cal U}}(X)/S^{\;{\cal V}}(A) @> >> S(X)/S(A), \end{CD}$$

resulting in a commutative diagram of chain complexes

$$\begin{CD} 0@> >> S^{\;{\cal V}}(A) @> >> S^{\;{\cal U}}(X) @> >>... ...} V @VV j V @.\\ 0@> >> S(A) @> >> S(X) @> >> S(X)/S(A) @> >> 0 \\ \end{CD}$$

Since the long exact sequence in homology is natural (exercise 6 of lecture 29), we get the commutative diagram:

$\begin{CD} @> >> H^{\;{\cal V}}_n(A) @> {i} >> H^{\;{\cal U}}_n(X) @> p >> H_n... ...(X) @> {p}>> H_n(S(X)/S(A)) @> >> H_{n-1}(A) @> >> H_{n-1}(X) @> >>\\ \end{CD}$

where we the subscript star indicates the map induced in homology. The five lemma enables us to conclude that

$$j_*\;:\; H_n(S^{\;{\cal U}}(X)/S^{\;{\cal V}}(A)) \longrightarrow H_n(S(X)/S(A)) = H_n(X, A).$$

is an isomorphism. Note the inclusion

$$k\;:\; S(X-U)\longrightarrow S^{\;{\cal U}}(X)$$

maps $S(A-U)$ into $S^{\;{\cal V}}(A)$ whereby we get an isomorphism (exercise 2)

$$\overline{k}\;:\;S(X-U)/S(A-U) = S^{\;{\cal U}}(X)/S^{\;{\cal V}}(A). \eqno(38.2)$$

The composite $j\circ \overline{k}$ is also induced by the inclusion map $(X-U,\; A-U)\longrightarrow (X, A)$ and we have the desired isomorphism

$$(j\circ\overline{k})_*\;:\; H_n(X-U, A-U)\longrightarrow H_n(X, A),\quad n = 0, 1, 2, \dots$$