Tensor algebra question 1)Why embedding of ( not necessarily finite-dimensional) vector spaces $V\rightarrow W$ produces embedding of tensor algebras $T(V)\rightarrow T(W)$. 
I can prove it using Hamel basis in $W$ but is there a nicer ( more functorial ) argument?
2) How to prove the same statement for modules over an algebra instead of vector spaces?
 A: If $V$ is a subspace of $W$, consider the inclusion $f:V\to W$ and any map $g:W\to V$ such that $g\circ f=1_V$; to construct $g$, you need to use bases or something equivalent, for it does not exist over, say, a general ring...
Now $T(-)$ is a functor, so $T(g)\circ T(f)=T(1_V)=1_{T(V)}$. It follows that the map $T(f):T(V)\to T(W)$ is injective.
A: Let me give another answer to 1).
In general, given a linear mapping $$\phi \colon E \to F$$ it extends uniquely to a homomorphism $$T(\phi) \colon T(E) \to T(F).$$ The proof can be made coordinate-free, in fact it follows from the universal property of $T(E)$ applied to the map $$\eta \colon E \to T(F),$$
where $\eta=i \circ \phi$ and $i \colon F \to T(F)$ is the natural embedding. 
By construction it follows
$$T(\phi)(x_1 \otimes \ldots \otimes x_p)=\phi x_1 \otimes \ldots \otimes \phi x_p.$$
If $\psi \colon F \to G$ is another linear map one obtains $$T(\psi \circ \phi)=T(\psi) \circ T(\phi),$$ hence $T(\phi)$ is injective [risp. surjective] whenever $\phi$ is injective [resp. surjective].
For more details, see for instance [Greub, Multilinear Algebra, Chapter III].
