I've proved by universal property that there exists $\Gamma: V \otimes W \to \mathcal{L}(V^{*},W)$ given by $$\Gamma(v \otimes w)(f)= f(v)w.$$
My question is, $\Gamma$ is injective? If $v \otimes w \in \ker \Gamma$, then $\Gamma(v \otimes w )(f) = 0$ for all $f \in V^{*}$. I mean, how can I proof that it implies $v\otimes w = 0?$
Building on mrtaurho's comment, the basic argument for FDVS is actually that this map is injective and then add the fact that it's finite dimensional so it's an isomorphism. Here's how you see it's injective.
If $\Gamma(v \otimes w)(f) = f(v)w = 0$ for all $f$, then either $w = 0$ or $f(v) = 0$ for all $f$. The former immediately implies that $v \otimes w = 0$, and as for the latter, since $f(v) = 0$ for all $f$, this is only true if $v = 0$ (why is this?), so again we have $v \otimes w = 0$. So the kernel is trivial, and the map is injective.