Suppose that $U$ is a cover of a topological space $X$ such that its interior too is a cover. So there is an isomorphism between the homology of $S_n^U(X)$ and that of $S_n(X)$.
In Vick’s algebraic topology book, it has been written that the hypothesis of interior of the cover being a cover is essential by bringing the following example: suppose $X=S^1$ the circle, $x_{0}\in X$ and $U=\{\{x_0\},X-\{x_0\}\}$. Any chain $c$ in $S_1^U(X)$ is uniquely written as the sum of a chain $c_1$ in $\{x_0\}$ and a chain $c_2$ in $X-\{x_0\}$. Since the image of $c_2$ is in a compact subset, $c$ will be a cycle if and only if both $c_1$ and $c_2$ are cycles. Also, then $c_1$ and $c_2$ must be boundaries.
My question is that where is compactness possibly used to show the claim? Also, how do we conclude that $c_1$ and $c_2$ are boundaries?
I don't know why they would mention compactness: if you proved the claim that any chain in $S_1^U(X)$ decomposes uniquely as $c_1+c_2$ for some $c_1\in S_1(\{x_0\}), c_2\in S_1(X-\{x_0\})$, then using this for $\partial c= \partial c_1 + \partial c_2$, then $\partial c = 0$ if and only if $(\partial c)_1 = 0 = (\partial c)_2$ and $(\partial c)_i= \partial c_i$ by uniqueness
About the fact that they're boundaries, it's pretty simple : $X\setminus \{x_0\}$ is homeomorphic to an interval, therefore it has trivial homology in degree $1$: a cycle is always a boundary; similarly $\{x_0\}$ has trivial homology in degree $1$ too (and actually in all degrees $>0$ for both)