Note that this an exercise in Foote and Dummit's Algebra.
Problem: Let $1\in R$. Any $R$-module $M$ is contained in an $R$-injective module.
Definition : $R$-module $Q$ is injective if $0\rightarrow L \rightarrow M, f: L \rightarrow Q$ has lift
Proof : I will follow the way in the book. For proof, I will enumerate facts :
(1) $Q$ is $R$-injective iff for any ideal $I$ with $g: I\rightarrow Q$ has extension $G: R\rightarrow Q$.
(2) ${\rm Hom}_{\bf Z} (R,M)$ is $R$-module.
(3) If $Q$ is $R$-injective then $ {\rm Hom}_{\bf Z} (R,Q)$ is $R$-injective.
(4) ${\rm Hom}_{R} (R,M)=M$
(5) Note that any ${\bf Z}$-module $M$ is contained in an ${\bf Z}$-injective module $Q$ and $$ (M=)\ {\rm Hom}_{\bf R} (R,M)\subset {\rm Hom}_{\bf Z} (R,M)\subset {\rm Hom}_{\bf Z} (R,Q)$$
So if ${\rm Hom}_{\bf Z} (R,Q)$ is $R$-injective we are done. At least it is ${\bf Z}$-injective. How can we finish the proof?
If we have R module homomorphism it has Z module lift So we need to show that it is R module homomorphism
In order to do this we shall use Baer's criterion (see Proposition 36(1) from the same book). Let $I\subset R$ be a left ideal and $f:I\to\operatorname{Hom}_{\mathbb Z}(R,Q)$ an $R$-linear mapping. It suffices to find $g\in\operatorname{Hom}_{\mathbb Z}(R,Q)$ such that $f(a)=ag$ for all $a\in I$. If $a\in I$ then $f(a)\in\operatorname{Hom}_{\mathbb Z}(R,Q)$ and $f(a)(x)\in Q$ for any $x\in R$. It follows that the map $I\ni a\mapsto f(a)(1)\in Q$ is $\mathbb Z$-linear, and since $Q$ is an injective $\mathbb Z$-module it can be extended to a map $h\in\operatorname{Hom}_{\mathbb Z}(R,Q)$. If $a\in I$ and $x\in R$, then $ah(x)=h(ax)=f(ax)(1)=xf(a)(1)=f(a)(x)$, so $f(a)=ah$ and thus $f$ can be extended to $R$ in an obvious manner.