Space with prescribed local homology 
Lets be $G_n$ sequence of abelian groups and $G_0 = \mathbb{Z}$. 
Is there topological space $X$ that local homology groups at every point are those $G_n$ ? ie. $$
\forall x\in X \; \forall n\in \mathbb{N}_0 \; : \: H_n(X,X\setminus \{x\}) = G_n
$$
Extra question: Can $X$ be embedded in Euclidean space? Or can $X$ be compact? If no, why?

It is not hard to find space $X$ that its homology groups are $G_n$ ie $H_n(X) = G_n$. 
With this space we can make space which has one point with prescribed local homology $G_n$. Let $\tilde{Y} = X \times [0,1]$. Define equivalence on $\tilde{Y}$ that $\{x,0\} \sim \{y,0\}$ for all $x,y \in X$. Now let $Y$ be factor space $Y = \tilde{Y}\big/_\sim$.  
Than I think $H_n(Y,Y\setminus [\{x,0\}]) = G_n$. I do not have proof of this yet, but I think it should work.
 A: There are two subquestions in your question:


*

*$H_n(Y, Y')\cong G_n$. Where $Y'= Y\setminus \{[x,0]\}$. Indeed, since $Y$ is contractible, we have the long exact sequence of reduced homology groups: 
$$
0= \tilde{H}_k(Y)\to H_k(Y, Y') \to \tilde{H}_{k-1}(Y') \to \tilde{H}_{k-1}(Y)=0 ...   
$$ 
Since 
$$
\tilde{H}_{k-1}(Y')\cong \tilde{H}_{k-1}(X)\cong G_{k-1}, 
$$
this means that you almost got it right:
$$
\tilde{H}_n(Y,Y')\cong G_{n-1} 
$$
except for $n=1$, where you are off by the factor of ${\mathbb Z}$:
$$
\tilde{H}_1(Y, Y')\cong G_0/{\mathbb Z}. 
$$ 

*The spaces $X$ with prescribed homology $G_n$ can be obtained as follows: You start with the $G_n$-Moore space $X_n=M(G_n,n)$, i.e., a space with $\tilde{H}_k(X_n)=0, k\ne n$ and $H_n(X_n)\cong G_n$. Then you take as $X$ the bouquet of the spaces $X_n$, $n\ge 0$.  

*Here is a sketch of a construction answering your other question (a space with prescribed local $n$-th homology $\cong G_{n-1}$ at every point). However, I did not check all the details, specifically, organizing the transfinite induction part:
Start with the space $Y$ as in Part 1. 
Step 1. At every point $z\in Y$ of $Y$, except for the "tip"  $y_0:= [x,0]$, attach a  copy $Y_z$ of $Y$. Denote this space by $Y_1$. It is not hard to check  that at every $y\in Y\subset Y_1$, we have
$$
H_n(Y_1, Y_1\setminus \{y\})\cong H_n(Y, Y\setminus \{y\})\cong G_{n-1}.  
$$
(To prove the first isomorphism use contractibility of $Y$: The spaces $Y_z$ that we are attaching are all contractible and disjoint from $y$.)
Now, repeat Step 1 at every $z\in Y_1\setminus Y$ (attach a copy of $Y$ at every such $y$). Thus, we obtain $Y_2$. Next, obtain $Y_3$ by attaching copies $Y_z$ of $Y$ at all points $z\in Y_2\setminus Y_1$. Continue this process via transfinite induction of length bounded by the cardinality of $Y$. Note that parts of the transfinite induction will involve taking direct limits. Here you use the fact that homology commutes with direct limits. 
