This is the construction
We define $S^m\times S^n\rightarrow S^{m+n+1}$ by:
$f(x,y)=(\frac{1}{\sqrt{2}}x,\frac{1}{\sqrt{2}}y)$
where $x\in S^m$ and $y\in S^n$. This function is well-defined. Since the components of $f$ are polynomial functions, and polynomial functions are continuous, then $f$ is continuous. Also, $f$ is injective as $f(x_1,y_1)=f(x_2,y_2)$ gives $x_1=x_2$ and $y_1=y_2$. That is $f$ maps distinct points in $S^m\times S^n$ to unique points in $S^{m+n+1}$.
Notice that $S^m$ and $S^n$ are compact. Therefore, their product, $S^m\times S^n$ is compact. Also, notice that $S^{m+n+1}\subseteq \mathbb{R}^{n+m+2}$ is a submanifold. And since $\mathbb{R}^{n+m+2}$ is Hausdorff, then $S^{n+m+1}$ is Hausdorff. Therefore, $f$ is a continuous injective map between a compact space and a Hausdorff space. Hence, $f$ is a homeomorphism.
Lastly, for $p=(x,y)\in S^m\times S^n$ given by $D_pf$ has $\frac{1}{\sqrt{2}}$ in its diagonal entries, and zeros elsewhere. This means the rows are linearly independent. Hence, the matrix has full rank (it has a rank of $n+m+2$). This matches the dimension of the target space $S^{n+m+1}$. Hence, $f$ is immersive at every point.
Since $f$ is a homeomorphism and an immersion, then we say it is an embedding.
I don't think there's much to improve about the construction itself apart from the issues pointed out in the comments. Personally, I would be a bit more concise since you spell out many rather basic facts that one could without problem assume known (such as that the spaces involved are compact Hausdorff). With that in mind, how about the following:
Let $f\colon S^n \times S^m \to S^{n + m + 1}$ be given by $f(x, y) = \frac{1}{\sqrt{2}}(x, y)$. This map is clearly well-defined and injective, hence a homeomorphism onto its image by the compact-Hausdorff lemma. Considering $f$ as a function $\newcommand{\R}{\mathbb{R}}\R^{n + 1} \times \R^{m + 1} = \R^{n + m + 2} \to \R^{n + m + 2}$, we have that $D f = \frac{1}{\sqrt{2}}\mathrm{Id}$ and thus $f$ is an immersion and altogether a (smooth) embedding.