Here is a fairly simple argument directly from the universal property. Given an abelian group $A$, define a group $F$ of "fractions" $\frac{a}{n}$ where $a\in A$ and $n\in\mathbb{Z}\setminus\{0\}$. We impose an equivalence relation on these fractions by saying $\frac{a}{n}=\frac{b}{m}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(ma-nb)=0$. Using the usual formula for addition of fractions, you can check that the set $F$ of equivalence classes of fractions forms an abelian group.
There is now a bilinear map $f:A\times\mathbb{Q}\to F$ defined by $f(a,m/n)=\frac{am}{n}$. This induces a homomorphism $g:A\otimes\mathbb{Q}\to F$ which sends $a\otimes 1$ to the fraction $\frac{a}{1}$. So to show that $a\otimes 1\neq 0$, it suffices to show the fraction $\frac{a}{1}$ is nonzero. The zero element of $F$ is the fraction $\frac{0}{1}$, and $\frac{a}{1}=\frac{0}{1}$ iff there exists $k\in\mathbb{Z}\setminus\{0\}$ such that $k(1\cdot a-1\cdot 0)=0$, or in other words such that $ka=0$. So if $a$ is a non-torsion element, then $a\otimes 1\neq 0$. (In fact, this map $g$ is actually an isomorphism, but that takes more work to prove.)
The case of $\mathbb{R}$ follows from the case of $\mathbb{Q}$ since $\mathbb{Q}$ is a direct summand of $\mathbb{R}$ and tensor products distribute over direct sums.
From a broader perspective, what's really going on here is that $\mathbb{Q}$ (or $\mathbb{R}$) is a flat $\mathbb{Z}$-module, meaning that if $A$ is a $\mathbb{Z}$-module and $B\subseteq A$ is a submodule, then the map $B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$ induced by the bilinear map sending $(b,q)\in\mathbb{Q}$ to $b\otimes q\in A\otimes\mathbb{Q}$ is injective. So, in particular, if $a\in A$ is any element such that $na\neq 0$ for all $n$, the submodule $B\subseteq A$ generated by $a$ is isomorphic to $\mathbb{Z}$. Since $\mathbb{Z}\otimes\mathbb{Q}\cong\mathbb{Q}$ and we have an injective map $\mathbb{Z}\otimes\mathbb{Q}\cong B\otimes\mathbb{Q}\to A\otimes\mathbb{Q}$, this implies $A\otimes\mathbb{Q}$ is nontrivial (and in fact $a\otimes1\neq 0$, since it is the image of the nonzero element $1\otimes 1\in \mathbb{Z}\otimes\mathbb{Q}$).
The argument using fractions above can be generalized to show that if $R$ is any commutative ring, then any localization of $R$ is flat as an $R$-module. For $\mathbb{Z}$-modules, flatness can be characterized quite simply in general: a $\mathbb{Z}$-module is flat iff it is torsion-free. See Show that a Z-module A is flat if and only if it is torsion free? for some sketches of the proof.
Best Answer
This can be again turned into question about the universal property: $\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}$ being zero would mean that any bilinear map with domain $\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ is zero. Therefore, it is e.g. sufficient to find a bilinear map $$\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z} \rightarrow \mathbb{Z}/2\mathbb{Z}$$ that is nonzero. You can try to find one. Similarly, in case you are interested in the actual elementary tensor $1 \otimes [1]$ being nonzero, you should find a bilinear map coming from $\mathbb{Z}\times \mathbb{Z}/2\mathbb{Z}$ that does not send $(1, [1])$ to $0$.
Alternatively: You noted before that for any ring $R$ and any $R$-module $M$, $R \otimes_R M \simeq M$. Thus, in our case ($R=\mathbb{Z}, M=\mathbb{Z}/2\mathbb{Z}$), one gets $\mathbb{Z}\otimes \mathbb{Z}/2\mathbb{Z} \simeq \mathbb{Z}/2\mathbb{Z}.$
And no, in general, the tensor product of two nonzero modules may be $0$. You can try to show, using the bilinear relations, that the module $\mathbb{Z}/3\mathbb{Z} \otimes \mathbb{Z}/2\mathbb{Z}$ is trivial.