Denote by $X^*$ the algebraic dual of a vector space $X$. Let $B(X\times Y)$ be the bilinear forms.
Define $X^* \times Y^* \to B(X\times Y),\quad (\alpha ,\beta)\mapsto B_{\alpha,\beta}$, where $$B_{\alpha,\beta}(x,y) := \alpha (x)\beta (y). $$ By invoking the universal property of the tensor product we get a linear $$X^* \otimes Y^* \to B(X\times Y),\quad u:=\sum_{k=1}^n \alpha _k\otimes \beta _k \mapsto B_u. $$ In Ryan's Tensor products it suggested that this map is injective, but I don't understand why. Suppose $$B_u(x,y) = \sum_{k=1}^n \alpha _k(x)\beta _k(y) = 0,\quad (x,y)\in X\times Y.\tag{1} $$ To show the corresponding tensor $u =0$ it would be equivalent to show (Prop. 1.2) $$\forall s\in X^{**} \forall t\in Y^{**}\ \sum_{k=1}^n s(\alpha _k)t(\beta _k) = 0 $$ or $$\forall s\in X^{**} \sum_{k=1}^n s(\alpha _k)\beta _k = 0 \qquad \mbox{or}\qquad \forall t\in Y^{**}\sum_{k=1}^n t(\beta _k)\alpha _k =0. $$ But I don't see how to make use of $(1)$. For instance, taking $s\in X^{**}$ we would have $$\left (\sum_{k=1}^n s(\alpha _k)\beta _k\right )(y) = \sum_{k=1}^n s(\alpha _k)\beta _k(y) \overset{?}= 0\qquad (y\in Y). $$ The above would hold in finite dimensions, because then $s$ is identified with $x^{**}$ for some $x$ and $x^{**}(\alpha _k) = \alpha _k(x)$ and $(1)$ applies.
The author doesn't explicitly assume finite dimension, but then I have no idea what $s(\alpha _k)$ is.
Does (1) imply $u=0$ for arbitrary vector spaces $X,Y$ and what is the justification?
Changing the $\beta_k's$ if necessary, you can assume without loss of generality that $\alpha_1,\ldots,\alpha_n$ are linearly independent (choose a maximal subset of independent $\alpha_k$'s, write the others as linear combinations of the elements in the subset, regroup terms, and use the bilinearity of the tensor product).
Fixing $y\in Y$, and letting $x$ range over $X$, $(1)$ show that you have a dependance relation $\displaystyle\sum_{k=1}^n\beta_k(y) \alpha_k=0$ in $X^*$. By assumption $\beta_k(y)=0$ for all $k$. But this is true for all $y\in Y$, so $\beta_k=0$ for all $k$ and $u=0$.