At the moment I know the fact for finitely generated abelian groups and this:
If $x \otimes y = 0 \in M \otimes N$, then there is a finitely generated submodule $L \subset M$ containing $x$ such that $x \otimes y = 0 \in L \otimes N$.
But I can't see how I can apply this. I have been trying to find a finitely generated submodule, but I can't find one that is applicable
On
A torsion abelian group is the direct limit of its finitely generated (torsion) subgroups and tensor product commutes with direct limits.
Some details
Write $A=\varinjlim\limits_{i\in I} A_i$ (direct limit over the finitely generated submodules of $A$, ordered by inclusion). Then $$A\otimes\mathbf Q= (\varinjlim\limits_{i\in I} A_i)\otimes\mathbf Q=\varinjlim\limits_{i\in I} (A_i\otimes\mathbf Q) =0.$$
You meant $A\otimes_{\mathbf{Z}} \mathbf{Q}$ I guess... Anyway.
As I am uncool, unfashioned and as I don't think that $\varinjlim$ solves anything here, I will give a simple proof.
If $A$ is torsion, take any simple tensor $a\otimes r \in A\otimes_{\mathbf{Z}} \mathbf{Q}$. As $A$ is torsion, there is an $n\in\mathbf{Z}\backslash\{0\}$ such that $na=0$, and then $a\otimes r = a\otimes \left( n\frac{r}{n}\right) = (na)\otimes \frac{r}{n} = 0 \otimes \frac{r}{n} = 0$. So any simple tensor is zero, and as simple tensors generate $A\otimes_{\mathbf{Z}} \mathbf{Q}$ as $\mathbf{Z}$-module, $A\otimes_{\mathbf{Z}} \mathbf{Q} = 0$. Inversely, suppose that $na\not=0$ for all $n\in\mathbf{Z}$. Then $\mathbf{Z} a$ is a free $\mathbf{Z}$-module of rank $1$, and then $\mathbf{Z} a\otimes_{\mathbf{Z}} \mathbf{Q}$ is a free $\mathbf{Q}$-module, i.e. a $\mathbf{Q}$-vector space, of dimension $1$, and since $a\otimes 1 \not= 0$ in $\mathbf{Z} a\otimes_{\mathbf{Z}} \mathbf{Q}$, there exists a $\mathbf{Z}$-bilinear map $\mathbf{Z} a \otimes_{\mathbf{Z}} \to P$, where $P$ is some $\mathbf{Z}$-module, that does not kill the element $(a,1)$, and as an extension $A\otimes_{\mathbf{Z}} \mathbf{Q}\to P$ of this bilinear map exists, one sees that $a\otimes 1\not=0$.
Remark 1. Recall that an element $m\otimes n\in M\otimes_k N$ is zero if and only if $(m,n)$ is sent to $0$ by any bilinear application defined on $M\times N$.
Remark 2. The best way to picture this is the following. Fix $A$ a commutative ring and a multiplicative part $S$ of $A$, you can construct a ring $S^{-1}A$ having the following property : this ring is the "smallest" $A$-algebra where elements of $S$ are invertible. Each element of $S^{-1}A$ can be written in the form $\frac{a}{s}$ with $a\in A$ and $s\in S$, and two elements $\frac{a}{s}$ and $\frac{a'}{s'}$ are equal if and only if there exists an $s''\in S$ such that $s''(s'a-sa')=0$. Namely, you construct $S^{-1}A$ has the quotient of the set $S\times A$ by the following equivalence relation : $(s,a)$ and $(s',a')$ are equivalent if and only if there exists an $s''\in S$ such that $s''(s'a-sa')=0$, and the equivalence class of $(s,a)$ is noted $\frac{a}{s}$, and you have a canonical morphism $i_S : A\to S^{-1}A$ defined by $a\to\frac{a}{1}$. Now, the universal property is the following : for any ring morphism $f : A \to B$ such that $f(S)\subseteq B^{\times}$ there exists a unique ring morphism $g : S^{-1}A\to B$ such that $g\circ i_S = f$, and $g$ is defined by $g\left( \frac{a}{s} \right) = f(a)f(s)^{-1}$. The ring $S^{-1}A$ is called the localization of $A$ with respect to $A$. Now, if $M$ is an $A$-module, you can look for a $S^{-1}A$-module $M'$ endowed with an $A$-linear map $j_S : M\to M'$ having the following universal property : for each $A$-module $Q$ and any $A$-linear map $F : M \to Q$ such that hometheties of ratios in $S$ are isomomorphism of $M'$, there exists a unique $A$-linear map $G : M' \to Q$ such that $G\circ j_S = F$. Such an $S^{-1}A$-module is note $S^{-1}M$ and you can construct in the same way as $S^{-1}A$ was constructed. The $S^{-1}A$-module $S^{-1}M$ is called the localization of $M$ with respect to the multiplicative subset $S$. Now, you can show that $S^{-1}M \simeq M\otimes_A S^{-1}A$. Now if an element $m\otimes 1 = \frac{m}{1}$ is zero, that is equal to $\frac{0}{1}$, by definition of the equivalence relation, it means that there exists an $s\in S$ such that $sm = 0$. To get you case, take $A = \mathbf{Z}$, $M = $ your abelian group, and $S = \mathbf{Z}\backslash\{0\}$. You therefore have $S^{-1}A = \mathbf{Q}$.