A complex unital algebra which is a Banach space is also a Banach algebra In my functional analysis class I got stuck on this question on Banach algebras:

Let $ \mathbb{A} $ be a complex unital algebra (it has a unit i.e. a multiplicative identity) and $ \mathbb{A} $ is equipped with a norm such that A is also a Banach space and with respect to this norm metric the multiplication operation is continuous in two components (meaning if for two sequences in $ \mathbb{A} $ such that $ \{ a_n \}_{n=1}^{\infty} \to a \in \mathbb{A} $ and $ \{ b_n \}_{n=1}^{\infty} \to b \in \mathbb{A} $ we have $ a_nb \to ab $ and $ ab_n \to ab $). We are asked to prove there is an equivalent norm on $ \mathbb{A} $ such that it is a Banach Algebra. We are also asked to look at the fact that the algebra has a unit and we are asked to see if this conclusion is necessarily true if the Algebra is not unital.

I cannot really seem to prove this or take advantage of the hint and could not find a reference for this in the literature so I am posting here in the hopes of finding help. Thanks all helpers.
 A: Let's define for $a\in A$ the left multiplication operator and the right multiplication operator
$$ L_a : A \rightarrow A, L_a(x)=ax \quad \text{and} \quad R_a: A \rightarrow A, R_a(x)=xa.$$
As multiplication is assumed to be continuous we get that $L_a$ and $R_a$ are both continuous for all $a\in A$. Let $B=B_1(0,A)$ denote the unit ball in $A$. Then holds for $x\in A$
$$ \sup\{ \Vert L_a(x) \Vert : a\in B \} = \sup \{ \Vert R_x(a) \Vert : a\in B \} = \Vert R_x\Vert < \infty.$$
The uniform boundedness principle tells us 
$$ C:=\sup\{ \Vert L_a\Vert : a\in B \} < \infty.$$
Let now $0\neq x\in A$ and $y\in A$ then
$$ \Vert xy\Vert= \Vert x \Vert \cdot \Vert \frac{x}{\Vert x \Vert} y\Vert
=\Vert x \Vert \cdot \Vert L_{\frac{x}{\Vert x \Vert}}(y) \Vert \leq \Vert x \Vert \cdot \Vert L_{\frac{x}{\Vert x \Vert}}\Vert \cdot \Vert y \Vert \leq C \Vert x \Vert \cdot \Vert y \Vert. $$
We define a new norm
$$ \Vert x \Vert_{1} := C \Vert x \Vert.$$
Then
$$\Vert xy\Vert_1= C \Vert x y \Vert \leq C^2 \Vert x \Vert \cdot \Vert y \Vert = \Vert x \Vert_1 \cdot \Vert y \Vert_1.$$
Hence, $A$ with this norm is a Banach algebra.
