In [1] it is said that the category $\mathbf{FinVect}$ of finite vector spaces is a tensor category. I am trying to convince myself that this is indeed the case.
One of the properties that $\mathbf{FinVect}$ must satisfy for this to hold is that it is indecomposable. I cannot think of a way to show that. Here are the definitions which are relevant for this problem:
Def: Let $\mathbb C$ be a locally finite $\mathbb{K}$-linear abelian rigid monoidal category. $\mathbb{C}$ is a multitensor category over $\mathbb{K}$ if $\otimes:\mathbb{C}\times\mathbb{C}\rightarrow \mathbb{C}$ is bilinear on morphisms. A multitensor category $\mathbb{C}$ is decomposable if it is equivalent to a direct sum of nonzero multitensor categories. If $\mathbb{C}$ is an indecomposable multitensor category and $\text{End}(1)\cong \mathbb{K}$ (as vector spaces), then $\mathbb{C}$ is a tensor category.
[1] Pavel Etingof et al. Tensor Categories. American Mathematical Society, 2015.
If $\mathbf{FinVect}\simeq C\times D$ where $C$ and $D$ are both nonzero, let $c$ and $d$ be nonzero objects of $C$ and $D$, respectively. Then $(c,0)$ and $(0,d)$ are both nonzero objects of $C\times D$ (because their identity map is not equal to their zero map), but $\operatorname{Hom}((c,0),(0,d))\cong \operatorname{Hom}(c,0)\times\operatorname{Hom}(0,d)\cong 0$. This is a contradiction, since any two nonzero vector spaces have a nonzero linear map between them.
For some similar arguments for other categories, see my answer at Is Set "prime" with respect to the cartesian product? and its comments.