We have a long exact sequence
$...\to H_0(A)\to H_0(X)\to H_0(X,A)\to 0$
If $H_0(X,A)=0$, then by exactness at $H_0(X)$: Image $(H_0(A)\to H_0(X))=H_0(X)=$ ker $(H_0(X)\to H_0(X,A))$ so the map $i_*:H_0(A)\to H_0(X)$ is surjective.
Let $X_a$ be a path component of $X$. Suppose on the contrary that $A\cap X_a=\emptyset$. I don't understand how to get contradiction from here and how to prove the converse. Please help. Thanks.
The linked post does not answer my question. The linked post does not clarify why A intersects every path component of $X$. The answer here Path-components and relative homology does not explain the steps so I don't understand it.
I don't understand the answer here either, it seems too complicated $H_0(X,A) = 0 \iff A$ meets each path-component of $X$. and same for this $H_0(X,A) = 0 \iff A \cap X_i \neq \emptyset \forall $ path-components $X_i$.
None of the linked posts explains how to use surjectivity of $i_*$ to show that each path component of X intersects $A$.