Is the cartesian product of two Hilbert spaces a Hilbert space? Let $(H_1,\langle , \rangle_1)$ and $H_2,\langle , \rangle_2)$ be two complex Hilbert spaces.

Is $H_1\times H_2$ a Hilbert space? 

I think that we can define the following inner-product on $H_1\times H_2$:
$$\langle X, Y\rangle=\langle x_1, y_1\rangle_1+\langle x_2, y_2\rangle_2,$$
for all $X=(x_1,x_2)\in H_1\times H_2$ and $Y=(y_1,y_2)\in H_1\times H_2$.

Question: why $(H_1\times H_2,\langle \cdot, \cdot\rangle)$ is complete?

 A: Hint the norm on $H_1\times H_2$ is given by
$$\|(x_1,x_2)\|_{H_1\times H_2} =\sqrt{\|x_1\|^2_{H_1}+\|x_2\|^2_{H_2}}$$
and under this norm it is easy to check the completeness of $H_1\times H_2$ since having a cauchy sequence $(x^n_1,x^n_2)_n$ in $H_1\times H_2$ will directly implies that $(x^n_i)_n$ is a Cauchy sequence(Why?) in the Hilbert (Complete space ) space $(H_i, \|\cdot\|_{H_i})$ $i=1,2$
because
$$\|x^n_i- x^m_i\|_{H_i}\le \|(x^n_1,x^n_2)-(x^m_1,x^m_2)\|_{H_1\times H_2} \to 0$$
can you proceed form here?
A: Consider the following inner product on $H_1\times H_2$ $\langle (x_1,x_2), (y_1,y_2)\rangle = \langle x_1,y_1 \rangle + \langle x_2, y_2\rangle$. It remains to show this is indeed an inner product.
Linearity
$$
        \langle a(x_1,x_2) + b(y_1,y_2), (z_1,z_2) \rangle \\
         =\langle (ax_1 + by_1, ax_2 + by_2), (z_1,z_2) \rangle\\
         = \langle ax_1 + by_1, z_1\rangle + \langle ax_2 + by_2, z_2\rangle\\
         = \langle ax_1 , z_1 \rangle  +\langle by_1, z_1\rangle + \langle ax_2 ,z_2 \rangle + \langle by_2, z_2\rangle\\
         =  a\langle(x_1,x_2), (z_1,z_2) \rangle + b\langle(y_1,y_2), (z_1,z_2) \rangle
    $$
Symmetry
$$\langle (x_1,x_2), (y_1,y_2)\rangle= \langle x_1, y_1 \rangle + \langle x_2, y_2 \rangle = \langle (y_1,y_2), (x_1,x_2)\rangle$$
Bilinearity
Assume $\langle(x_1,x_2), (x_1,x_2) \rangle>0$ then $$\langle(x_1,x_2), (x_1,x_2) \rangle= \langle x_1,x_1\rangle + \langle x_2,x_2 \rangle > 0 \implies $$ either $\langle x_1, x_1 \rangle > 0 $ or $\langle x_2, x_2 \rangle > 0 $ so $(x_1,x_2) \neq (0,0)$
Fix some Cauchy sequence in $\{ \left(x_{n}, y_{n}\right)\}  \in H_{1} \times H_{2}$. Then $x_n$ and $y_n$ are both Cauchy sequences in $H_1$ and $H_2$ and hence converge to $x$ and $y$ respectively. Now $(x,y) \in H_1 \times H_2$ and we claim this is the limit of $\{ \left(x_{n}, y_{n}\right)\}$.
\begin{align*}
&\left\|\left(x_{n}, y_{n}\right)-(x, y)\right\|_{H_{1} \times H_{2}} \\
=&\left\langle\left(x_{n}, y_{n}\right)-(x, y),\left(x_{n}, y_{n}\right)-(x, y)\right\rangle^{1 / 2} \\
=&\left\langle\left(x_{n}-x, y_{n}-y\right),\left(x_{n}-x, y_{n}-y\right)\right\rangle^{1 / 2} \\
=& \sqrt{\left\|x_{n}-x\right\|_{H_{1}}^{2}+\left\|y_{n}-y\right\|_{H_{2}}^{2}} \\
& \rightarrow 0
\end{align*}
as $x_{n} \rightarrow x$  and $y_{n} \rightarrow y$.
