I am looking at the co-unital property of a coalgebra.
This is my work:
Let $\epsilon: C\to \Bbb C$ be the counit, and let $\Delta$ be the coproduct of $C$, a coalgebra. So $(C,\Delta,\epsilon)$ is a coalgebra.
To satisfy counit requirements, we need:
$$(\epsilon \otimes \text{Id})(\Delta(c))=(\text{Id}\otimes \epsilon)(\Delta(c))$$
$$(\epsilon \otimes \text{Id})\left(\sum c_{(1)}\otimes c_{(2)}\right)=(\text{Id}\otimes \epsilon)\left(\sum c_{(1)}\otimes c_{(2)}\right)$$ And from what I understand, this is then: $$\sum\epsilon(c_{(1)})\otimes c_{(2)} = \sum c_{(1)} \otimes \epsilon(c_{(2)})$$
They write:
$$\sum\epsilon(c_{(1)})c_{(2)} = \sum c_{(1)}\epsilon(c_{(2)})$$
Why is this not a tensor product? Is this because it lives in $\Bbb C\otimes C$ and since this means that $\epsilon(\cdots)$ is just a scalar, that $\Bbb C\otimes C$ is isomorphic to..
Yes, your guess is exactly right. There is a canonical isomorphism $\mathbb{C}\otimes C\cong C$ which sends $a\otimes c$ to $ac$, and they are identifying $\mathbb{C}\otimes C$ with $C$ via this isomorphism (and similarly identifying $C\otimes\mathbb{C}$ with $C$ on the right-hand side of the equation). Note that you must make some such identification for the counit property to even make sense: in the identity $\sum\epsilon(c_{(1)})\otimes c_{(2)} = \sum c_{(1)} \otimes \epsilon(c_{(2)})$ that you write, the left-hand side is an element of $\mathbb{C}\otimes C$ and the right-hand side is an element of $C\otimes \mathbb{C}$. So for it to be meaningful for the two sides to be equal, you must be choosing some identification of $\mathbb{C}\otimes C$ with $C\otimes\mathbb{C}$ (and the right one to choose is the one that identifies both of them with $C$ as above).