For simplicity we work over a fixed algebraically closed field $\mathbb{k}$ of char 0. Recall that a linear algebraic group $G$ is a closed subgroup of $GL(n,\mathbb{k})$, and we say $G$ is linearly reductive if $Rep(G)$ is semi-simple, i.e. all representations decompose into a direct sum of irreducible representations.
My question is: given two linealy reductive algebraic groups $G$ and $H$, how to show that $G\times H$ is also linearly reductive?
Let's consider finite dimensional rational representations only.
An algebraic group $G$ is said to be linearly reductive if its "rational" or algebraic representations are semisimple.
Suppose that $G$ and $H$ are linearly reductive, and let $(V,\rho)$ be a finite-dimensional representation of $G\times H$. (Algebraic representations are locally finite, so there is no loss of generality in assuming $V$ is finite dimensional). Then, as a $G$-representation, $V$ decomposes canonically into a direct sum of isotypic summands, say $V = \bigoplus_{\lambda \in P_V} V_{\lambda}$ indexed by the irreducible representations of $G$ which occur in $V$, which we have labelled $\lambda$ where $\lambda$ runs over a finite set of isomorphism classes of irreducible $G$-representation, and $V_\lambda := \sum_{[U]=\lambda} U$ is defined to be the sum of all subrepresentations of $V$ which lie in the isomorphism class $\lambda$ (thus we write $[U]$ for the isomorphism class of $U$ as a $G$-representation).
Now the action of $H$ commutes with that of $G$, thus if $h \in H$ then $\rho(h) \in \mathrm{Hom}_G(V,V)$. But if $W$ is an irreducible subrepresentation of $V$, by Schur's Lemma any $G$-homomorphism must either act by $0$ on $W$ or map it to a subrepresentation isomorphic to $W$. Hence the $V_\lambda$ are preserved by $H$.
Let $U_{\lambda}$ be an irreducible $G$-representation in the isomorphism class of $G$-representations given by $\lambda$. Then we have a natural map $$ \mathrm{ev}\colon \text{Hom}_G(U_\lambda,V_{\lambda})\otimes U_\lambda \to V_\lambda, \quad \phi\otimes u \mapsto \phi(u). $$ Now $V_\lambda$ is completely reducible as a $G$-representation because $G$ is linearly reductive, so we may write it as a direct sum of irreducible representations, $V_{\lambda} = W_1\oplus W_2\oplus\ldots \oplus W_l$ where each $W_i$ is irreducible and so, by the definition of $V_{\lambda}$ the isomorphism class of each $W_i$ is $\lambda$. But then if $U_\lambda$ is any irreducible $G$-representation in the isomorphism class of $\lambda$ it follows that $$ \begin{split} \mathrm{Hom}_G(U_\lambda,V_{\lambda}) &= \mathrm{Hom}_G(U_\lambda,\bigoplus_{1\leq i\leq l} W_i) \\ &= \bigoplus_{1\leq i \leq l} \mathrm{Hom}_G(U_\lambda,W_i). \end{split} $$ Now by Schur's Lemma again, $\mathrm{Hom}_G(U_\lambda, W_i)$ is $1$-dimensional for each $i$ ($1\leq i \leq l$). Thus if we pick, for each $i \in \{1,2,\ldots,l\}$, $\phi_i \in \mathrm{Hom}(U_\lambda,W_i)$, $\phi_i \neq 0$, then $\{\phi_1,\ldots,\phi_l\}$ form a basis of $\mathrm{Hom}_G(U_\lambda,V_\lambda)$. Any element of $\mathrm{Hom}_G(U_\lambda,V_{\lambda})$ can therefore be written uniquely in the form $\sum_{i=1}^l \phi_i\otimes u_i$ where $u_i \in U_\lambda$. But if $v \in V_\lambda$, then $v= \sum_{i=1}^l w_i \in V_\lambda$, for unique $w_i\in W_i$ ($1\leq i \leq l$) and thus we may define $\psi(v) = \sum_{i=1}^l \phi_i\otimes \phi_i^{-1}(w_i)$. It follows that $\mathrm{ev}(\psi(v))=v$ and similarly $\psi(\mathrm{ev}(\sum_{i=1}^l \phi_i\otimes u_i)) = \psi(\sum_{i=1}^l \phi_i(u_i)) = \sum_{i=1}^l \phi_i\otimes u_i$, so that $\psi\circ \mathrm{ev} = \mathrm{id}$ and $\mathrm{ev}\circ \psi = \mathrm{id}$, i.e., $\mathrm{ev}$ is an isomorphism.
[Note that the decomposition of each $V_\lambda$ into a direct sum of irreducible subrepresentations, unlike the decomposition of $V$ into the isotypical subrepresentations $V_\lambda$, is not unique, and so it is not preserved by the action of $H$ (unless $V_\lambda$ is itself irreducible). ]
Now $H$ acts on $\mathrm{Hom}_G(U_\lambda,V_{\lambda})$ via $\rho(h)(\phi)(u) = h(\phi(u))$ thus it is a finite-dimensional $H$-representation. Since $H$ is linearly reductive, it therefore decomposes into a direct sum $\mathrm{Hom}_G(U_{\lambda},V_{\lambda}) = \bigoplus_{r=1}^s T_r$, where each $T_i$ is an irreducible $H$-subrepresentation. Now $\mathrm{Hom}_G(U_\lambda,V_{\lambda})\otimes U_{\lambda}$ is a $G\times H$-representation via $(g,h)(\phi\otimes u) = (\rho(h)\circ \phi)\otimes \rho(g)(u)$, for all $g\in G, h\in H, \phi \in \mathrm{Hom}_G(U_\lambda,V_\lambda)$, $u\in U_\lambda$, and it is easy to check that $\mathrm{ev}$ is a map of $G\times H$-representations. It follows that $V_{\lambda}$ decomposes into a direct sum of subrepresentations $V_\lambda = \bigoplus_{r=1}^s \mathrm{ev}(T_r\otimes U_{\lambda})$.
Thus it follows that $V_\lambda$, and hence $V$ is a direct sum of irreducible $G\times H$-representations once we show that $T_i\otimes U_\lambda$ is an irreducible $G\times H$ representation. But this actually follows from what we have already done: Suppose that $S$ is a nonzero subrepresentation of $T_i\otimes U_{\lambda}$. Then applying the above to $S$ we see that it must be a direct sum of subrepresentations of the form $\bigoplus_{j=1}^d B_j\otimes C_j$ where each $B_j$ is an irreducible $H$-representation and each $C_j$ is an irreducible $G$-representation. Moreover since $T_i\otimes U_{\lambda}$, as a $G$-representation, is just $\dim(T_i)$ copies of $U_\lambda$, all of the $C_j$s must be isomorphic to $U_\lambda$. Switching the roles of $G$ and $H$ the shows that all of the $B_j$s must be isomorphic to $T_i$, and then $\dim(B_j\otimes C_j) = \dim(T_i\otimes U_\lambda)$ and so we must have $d=1$ and $S = T_i\otimes U_\lambda$. Thus $T_i\otimes U_{\lambda}$ is irreducible as required.