Prove two equation about $H_{p} (X) $ and $kernel (i_{n_{*}})_{p} $ .

for the first statement we can say, $Delta^{p}$ s are closed and bounded so are compact and since $sigma :Delta^{p} rightarrow X$ are continous maps then It induced a compact space on X so Image of the $sigma$ is also compact.

by e theorem a compact subspace (here $Delta^{p}$) of a space (here X ) that bigger have a covering the subspace have a finite covering {if I am using a right theorem!}.

thus, Image $sigma subseteq bigcup^{m}_{k=0} U_{k}$. meaning that finite number of $U_{k}$s cover the Image $sigma$ since the $U_{n}$s are nested we could find the one k for $U_{k}$ and bring up all $U_{i}$ from 0 to k.

now for $S_{p}(X) = {{sigma | sigma: Delta^{p}rightarrow X }}$ we have:

$hspace{4 cm} S_{p}(X)=S_{p}(bigcup_{0}^{k} U_{i})= S_{p}(U_{k})$

and coresponding to boundary map $partial_{p}: S_{p}(X)rightarrow S_{p-1}(X)$ we can rewrite it to:

$hspace{4 cm}partial_{p}:S_{p}(U_{k})rightarrow S_{p-1}(U_{k})$

remember; $(i_{n*})_p: H_{p}(U_{n}) rightarrow H_{p}(X)$

can say; $bigcup_{0}^{infty} Image(i_{n*})_p = Image bigcup_{0}^{infty} H_{p}(U_{n}) rightarrow H_{p}(X)$

we can show that :

$hspace{4 cm}Image (i_{n*})_{p} subseteq H_{p}(X)$

and so:

$forall nin N;hspace{4 cm} Image (i_{n*})_{p} subseteq H_{p}(U_{k})$

then:

$hspace{4 cm}bigcup ^{infty} _{0} Image (i_{n*})_{p}subseteq H_{p}(X)$

and because:

$hspace{4 cm} H_{p}(U_{k})subseteq bigcup ^{infty} _{0} Image (i_{n*})_{p}$

then:

$hspace{4 cm} H_{p}(X) subseteq bigcup ^{infty} _{0} Image (i_{n*})_{p}$

thus:

$hspace{4 cm} H_{p}(X) = bigcup ^{infty} _{0} Image (i_{n*})_{p}$

edit: I think my former answer is clearer, but anyway I try to write the answer as detailed as possible.

Using the compactness of $Delta^p$, the image of any singular simplex $Delta^pto X$ (which is also compact) is contained in some $U_n$. Since any $p$-cycle $sigma$ of $X$ is a finite formal sum $sum_i sigma_i$ of singular simplices $sigma_i: Delta^p to X$, we can find $n$ such that $U_n$ contains the image of all $sigma_i$'s (by taking the maximum of $n_i$'s such that the image of $sigma_i$ is contained in $U_{n_i}$.) This means that there exists $sigma_i': Delta^pto U_n$ such that $i_ncirc sigma_i' = sigma_i$. Define $sigma' = sum_isigma_i'$, so that $(i_n)_ast sigma'=sigma$.

Now let $[sigma]in H_p(X)$ be the homology class represented by the $p$-cycle $sigma$. To prove the first statement, it suffices to prove that $[sigma]$ is in the image of $H_p(U_n)to H_{p}(X)$. This is true because $(i_n)_ast [sigma'] = [sigma]$.

For the second, let $sigma=sum_i{sigma_i}$ (where $sigma_i$'s are singular $p$-simplices) be a $p$-cycle in $U_n$ and suppose $[sigma]in H_p(U_n)$ satisfies $(i_n)_ast [sigma] = 0$. By definition this means that there exists a $(p+1)$-cycle $tau = sum_j tau_j$ in $X$ such that $partial tau =(i_n)_astsigma$ as $p$-cycles of $X$. By exactly the same argument as above, there exists $mgeq n$ such that $U_m$ contains all the images of $tau_j$'s. Now define $(p+1)$-simplices $tau'_j: Delta^{p+1}to U_m$ and a $(p+1)$-cycles $tau' = sum_j tau'_j$ in $U_m$ as above. Obsereve that $(i_m)_astpartial tau' = partial (i_m)_ast tau' = partial tau = (i_n)_ast sigma = (i_m)_ast(i_{m, n})_ast sigma$ as $p$-cycles of $X$. Since $(i_m)_ast: S_p(U_m)to S_p(X)$ is injective, this implies $partialtau' = (i_{m, n})_ast sigma$ as $p$-cycles in $U_m$. This means that, in $H_p(U_m)$, we have $(i_{m, n})_ast[sigma] = [partial tau']=0$, and thus any $[sigma]in ker((i_n)_ast)$ is contained in $ker((i_{m, n})_ast)$ for some $m$.