Is injectivity of algebras preserved by tensor products? 
Suppose $R' \subset R$, $S'\subset S$ are inclusion of $k$-algebras. Does it hold that $R'\otimes_kS' \rightarrow R \otimes_k S$ is injective ?

I know there're counterexamples for modules, but why does the algebra make things different?
I found it used as a lemma in the Stacks project to prove Lemma 00I3.

No, $R'\otimes_kS' \rightarrow R \otimes_k S$ needn't be injective: here is a counterexample. 
Take $k=\mathbb Z$, $R'=\mathbb Z\hookrightarrow R=\prod_{n=1}^\infty  \mathbb Z/2^n \mathbb Z$ the unique morphism,  and $S'= \mathbb Z/3 \mathbb Z \stackrel {id}{\to} S=\mathbb Z/3 \mathbb Z $.  
Then the morphism $R'\otimes_k S'=\mathbb Z/3\mathbb Z \rightarrow R \otimes_k S=0$ is not injective because its source is non-zero and its target is zero.  
[The equality $ R \otimes_k S=R\otimes _\mathbb Z \mathbb Z/3\mathbb Z=R/3R=0$ follows from the fact that $3$ is invertible in $R=\prod_{n=1}^\infty  \mathbb Z/2^n \mathbb Z$ since it is invertible in all the factors $\mathbb Z/2^n \mathbb Z$]
