At Murphy's book about $C^*$-algebras I found this definition:
A representation of a $\mathrm{C}^*$-algebra $A$ is a pair $(H, \varphi)$ where $H$ is a Hilbert space and $\varphi: A \rightarrow B(H)$ is a $*$-homomorphism. We say $(H, \varphi)$ is faithful if $\varphi$ is injective.
If $\left(H_\lambda, \varphi_\lambda\right)_{\lambda \in \Lambda}$ is a family of representations of $A$, their direct sum is the representation $(H, \varphi)$ got by setting $H=\oplus_\lambda H_\lambda$, and $\varphi(a)\left(\left(x_\lambda\right)_\lambda\right)=$ $\left(\varphi_\lambda(a)\left(x_\lambda\right)\right)_\lambda$ for all $a \in A$ and all $\left(x_\lambda\right)_\lambda \in H$. It is readily verified that $(H, \varphi)$ is indeed a representation of $A$. If for each non-zero element $a \in A$ there is an index $\lambda$ such that $\varphi_\lambda(a) \neq 0$, then $(H, \varphi)$ is faithful.
But, or what I verified is wrong (below), or the "readily verified" is wrong. I believe that the definition of $H=\oplus_\lambda H_\lambda$ is the set of $(x_\lambda)_{\lambda}\in\prod_\lambda H_\lambda$ such that $\sup_\lambda\|x_\lambda\|<\infty$, if this is the case, it's easy to check that $H$ is Banach space with norm $\|(x_\lambda)_\lambda\|=\sup_\lambda\|x_\lambda\|$, but is not Hilbert space.
Indeed, Let $H_\lambda=\mathbb{C}$, $\lambda=1,2$, with $\langle x,y\rangle=xy$. Consider $H=\oplus_1^2\mathbb{C}=\mathbb{C}\times\mathbb{C}$. So, if I use $x=(1,0),y=(1,2)\in H$ and the parallelogram law we have \begin{align*} \|x+y\|^2+\|x-y\|^2 =\|(2,2)\|^2+\|(0,-2)\|^2 =2^2+2^2=8 \neq10 =2(1^2+2^2) =2(\|x\|^2+\|y\|^2). \end{align*}
So $H$ is not Hilbert space. It's correct?
Actually, $\bigoplus_\lambda H_\lambda$ is defined to be set of $(x_\lambda)_\lambda$ such that $\sum_\lambda \|x_\lambda\|^2<\infty$. Note that, in the event that the indexing set is uncountable, this entails that $x_\lambda=0$ for all but countably-many $\lambda$. The vector space operations are the usual entrywise ones and the inner product is defined by $\langle (x_\lambda)_\lambda, (y_\lambda)_\lambda \rangle = \sum_\lambda \langle x_\lambda,y_\lambda\rangle$.