Let $X,Y,U,V$ be vector spaces. Let $S:X\to U$ and $T:Y\to V$ be linear maps. Then by linearisation there exists a linear map $$S\otimes T : X\otimes Y \to U\otimes V,\quad x\otimes y \mapsto (Sx) \otimes (Ty). $$ If $S$ and $T$ are injective, then their tensor product is, too. I think I managed to justify this as follows.
Note that $U = S(X) \oplus U'$. Define $S' : U\to X$ such that $S'(S(x_u) +u' ) = x_u$. One readily verifies linearity and that is a left inverse for $S$. Analogously, define $T':V\to Y$. Then by linearisation, there exists linear $$S'\otimes T' : U\otimes V \to X\otimes Y,\quad (S(x_u)+u') \otimes (T(y_v) + v') \mapsto x_u\otimes y_v. $$ This would give a left inverse for $S\otimes T$ hence making it injective.
I seem to recall that tensoring injective morphisms between modules does not always preserve injectivity. In fact, a certain module $F$ for which $\mbox{id}_F \otimes \phi$ is injective for all injective morphisms $\phi$ is called flat.
What am I invoking here that allows this to work for vector spaces?
Is it the decomposition $U=S(X) \oplus U'$?
I know, this is an old question, but I both hate it when questions are solely answered within the comments, and I think the crucial term has not been named yet:
For a field $\mathbb{K}$, vector spaces over $\mathbb{K}$ are the same as free $\mathbb{K}$-modules! Freeness means exactly that they admit a basis, and it implies the flatness as $\mathbb{K}$-modules. Of course, the proof for this goes exactly as it was explained in the comments. It is an important and useful property to keep in mind.