Lemma: Let $K\subset X$ and $L\subset Y$ be compact and let $U\subset X\times Y$
be open with $K\times L\subset U$. Then there are open sets $V\subset X$
and $W\subset Y$ with
$$
K\times L\subset V\times W\subset U.
$$
Proof: For each $\left(x,y\right)\in K\times L$ there are -- by definition
of the product topology -- open sets $V_{x,y}\subset X$ and $W_{x,y}\subset Y$
with $\left(x,y\right)\in V_{x,y}\times W_{x,y}\subset U$.
By compactness of $K$, there is thus for each $y\in L$ some $n_{y}\in\mathbb{N}$
and suitable $x_{1}^{\left(y\right)},\dots,x_{n_{y}}^{\left(y\right)}\in X$
with
$$
K\subset\bigcup_{i=1}^{n_{y}}V_{x_{i}^{\left(y\right)},y}=:V_{y}.
$$
Note that
$$
W_{y}:=\bigcap_{i=1}^{n_{y}}W_{x_{i}^{\left(y\right)},y}\subset Y
$$
is open with $y\in W_{y}$. By compactness of $L$, there are thus
$y_{1},\dots,y_{m}\in L$ with $L\subset\bigcup_{j=1}^{m}W_{y_{j}}=:W$.
Set
$$
V:=\bigcap_{j=1}^{m}V_{y_{j}}
$$
and note that $V\subset X$ and $W\subset Y$ are open with $K\times L\subset V\times W$,
so that it remains to show $V\times W\subset U$.
To this end, let $\left(x,y\right)\in V\times W$ be arbitrary. By
definition of $W$, there is $j\in\left\{ 1,\dots,m\right\} $ with
$y\in W_{y_{j}}$. By definition of $V$, we have $x\in V_{y_{j}}=\bigcup_{i=1}^{n_{y}}V_{x_{i}^{\left(y_{j}\right)},y_{j}}$
and hence $x\in V_{x_{i}^{\left(y_{j}\right)},y_{j}}$ for some $i\in\left\{ 1,\dots,n_{y}\right\} $.
But by definition of $W_{y_{j}}$, we have $y\in W_{x_{i}^{\left(y_{j}\right)},y_{j}}$
and hence $\left(x,y\right)\in V_{x_{i}^{\left(y_{j}\right)},y_{j}}\times W_{x_{i}^{\left(y_{j}\right)},y_{j}}\subset U$. $\square$
Now, we can use the usual proof to show $\lambda\left(K\times L\right)=\mu\left(K\right)\nu\left(L\right)$
for all compact $K,L$. Let $\varepsilon\in\left(0,1\right)$ be arbitrary.
By regularity, there are open sets $U\subset X\times Y$ and $V_{1}\subset X$
as well as $W_{1}\subset Y$ with $K\times L\subset U$, $K\subset V_{1}$
and $L\subset W_{1}$ and with
\begin{align*}
\lambda\left(K\times L\right) & \leq\lambda\left(U\right)<\lambda\left(K\times L\right)+\varepsilon,\\
\mu\left(K\right) & \leq\mu\left(V_{1}\right)<\mu\left(K\right)+\frac{\varepsilon}{1+\nu\left(L\right)},\\
\nu\left(L\right) & \leq\nu\left(W_{1}\right)<\nu\left(L\right)+\frac{\varepsilon}{1+\mu\left(K\right)}.
\end{align*}
By the Lemma, there are open sets $V_{2}\subset X$ and $W_{2}\subset Y$
with $K\times L\subset V_{2}\times W_{2}\subset U$. Set $V_{3}:=V_{1}\cap V_{2}$
and $W_{3}:=W_{1}\cap W_{2}$ and note $K\times L\subset V_{3}\times W_{3}\subset U$.
By Uryshon's Lemma, there are $\varphi\in C_{c}\left(X\right)$ and
$\psi\in C_{c}\left(Y\right)$ with $0\leq\varphi\leq1$, $0\leq\psi\leq1$
and $\varphi\equiv1$ on $K$ as well as ${\rm supp}\,\varphi\subset V_{3}$.
Analogously, $\psi\equiv1$ on $L$ and ${\rm supp}\,\psi\subset W_{3}$.
Define $\varphi\otimes\psi\left(x,y\right):=\varphi\left(x\right)\psi\left(y\right)$.
Then
$$
\mu\left(K\right)\nu\left(L\right)\leq\int\varphi\,{\rm d}\mu\cdot\int\psi\,{\rm d}\nu\leq\mu\left(V_{3}\right)\nu\left(W_{3}\right)\leq\mu\left(V_{1}\right)\nu\left(W_{1}\right)<\mu\left(K\right)\nu\left(L\right)+3\varepsilon.
$$
Furthermore,
$$
\lambda\left(K\times L\right)\leq\int\varphi\otimes\psi\,{\rm d}\lambda\leq\lambda\left(V_{3}\times W_{3}\right)\leq\lambda\left(U\right)<\lambda\left(K\times L\right)+\varepsilon.
$$
But $\int\varphi\otimes\psi\,{\rm d}\lambda=\int\varphi\,{\rm d}\mu\cdot\int\psi\,{\rm d}\nu$,
which finally implies
$$
\left|\mu\left(K\right)\nu\left(L\right)-\lambda\left(K\times L\right)\right|\leq3\varepsilon+\left|\int\varphi\,{\rm d}\mu\cdot\int\psi\,{\rm d}\nu-\int\varphi\otimes\psi\,{\rm d}\lambda\right|+\varepsilon=4\varepsilon.
$$
Since $\varepsilon>0$ was arbitrary, we obtain the claim for compact
sets $K,L$.
Now, let $U\subset X$ and $V\subset Y$ be open and of finite measure.
Let $\alpha<\lambda\left(U\times V\right)$ be arbitrary. By inner
regularity of $\lambda$, there is a compact set $T\subset U\times V$
with $\alpha<\lambda\left(T\right)\leq\lambda\left(U\times V\right)$.
Let $\pi_{1}:X\times Y\to X$ be the usual projection and define $\pi_{2}$
analogously. Note that $K_{1}:=\pi_{1}\left(T\right)\subset U$ and
$L_{1}:=\pi_{2}\left(T\right)\subset V$ are compact with $T\subset K_{1}\times L_{1}\subset U\times V$.
Hence,
$$
\alpha<\lambda\left(T\right)\leq\lambda\left(K_{1}\times L_{1}\right)\leq\lambda\left(U\times V\right).
$$
By inner regularity of $\mu,\nu$, there are also compact sets $K_{2}\subset U$
and $L_{2}\subset V$ with
\begin{align*}
\mu\left(K_{2}\right) & \leq\mu\left(U\right)<\mu\left(K_{2}\right)+\varepsilon,\\
\nu\left(L_{2}\right) & \leq\nu\left(V\right)<\nu\left(L_{2}\right)+\varepsilon.
\end{align*}
Now, let $K_{3}:=K_{1}\cup K_{2}$ and $L_{3}:=L_{1}\cup L_{2}$.
We then have $K_{3}\times L_{3}\subset U\times V$ and
$$
\alpha<\lambda\left(K_{1}\times L_{1}\right)\leq\lambda\left(K_{3}\times L_{3}\right)\leq\lambda\left(U\times V\right)
$$
as well as
$$
\alpha<\lambda\left(K_{3}\times L_{3}\right)=\mu\left(K_{3}\right)\nu\left(L_{3}\right)\leq\mu\left(U\right)\nu\left(V\right).
$$
Since $\alpha<\lambda\left(U\times V\right)$ was arbitrary, we see
$\lambda\left(U\times V\right)\leq\mu\left(U\right)\nu\left(V\right)<\infty$.
Furthermore, we get
$$
\left(\mu\left(U\right)-\varepsilon\right)\left(\nu\left(V\right)-\varepsilon\right)\leq\mu\left(K_{2}\right)\nu\left(L_{2}\right)\leq\mu\left(K_{3}\right)\nu\left(L_{3}\right)=\lambda\left(K_{3}\times L_{3}\right)\leq\lambda\left(U\times V\right).
$$
By sending $\varepsilon\downarrow0$, we see $\mu\left(U\right)\nu\left(V\right)\leq\lambda\left(U\times V\right)$
and hence
$$
\lambda\left(U\times V\right)=\mu\left(U\right)\nu\left(V\right)
$$
for all open sets $U\subset X$ and $V\subset Y$ of finite measure.
Now, by $\sigma$-finiteness and outer regularity of $\mu,\nu$, we
can find increasing families $\left(U_{n}\right)_{n\in\mathbb{N}},\left(V_{n}\right)_{n\in\mathbb{N}}$
of open sets with $X=\bigcup_{n}U_{n}$ and $Y=\bigcup_{n}V_{n}$
as well as $\mu\left(U_{n}\right)<\infty$ and $\nu\left(V_{n}\right)<\infty$.
For $n\in\mathbb{N}$ and any open set $V\subset Y$ of finite measure,
$$
\mathcal{G}_{n,V}:=\left\{ M\subset X\,\mid\, M\text{ Borel and }\lambda\left(\left[M\cap U_{n}\right]\times V\right)=\mu\left(M\cap U_{n}\right)\nu\left(V\right)\right\} .
$$
By what we just showed, $\mathcal{G}_{n,V}$ contains the $\pi$-system
of all open sets. Furthermore, it is not hard to see that $\mathcal{G}_{n,V}$
is closed under disjoint countable unions. Finally, for $M\in\mathcal{G}_{n,V}$,
we have
\begin{align*}
\lambda\left(\left[M^{c}\cap U_{n}\right]\times V\right)+\mu\left(M\cap U_{n}\right)\nu\left(V\right) & =\lambda\left(\left[M^{c}\cap U_{n}\right]\times V\right)+\lambda\left(\left[M\cap U_{n}\right]\times V\right)\\
& =\lambda\left(U_{n}\times V\right)\\
& =\mu\left(U_{n}\right)\nu\left(V\right)\\
& =\mu\left(M\cap U_{n}\right)\nu\left(V\right)+\mu\left(M^{c}\cap U_{n}\right)\nu\left(V\right),
\end{align*}
so that subtraction (everything is finite) shows $\lambda\left(\left[M^{c}\cap U_{n}\right]\times V\right)=\mu\left(M^{c}\cap U_{n}\right)\nu\left(V\right)$
and hence $M^{c}\in\mathcal{G}_{n,V}$. Thus, $\mathcal{G}_{n,V}$
is a $\lambda$-system.
By Dynkin's $\pi$-$\lambda$-Theorem, we conclude $\mathcal{G}_{n,V}=\mathcal{B}\left(X\right)$,
the $\sigma$-Algebra of all Borel sets. By continuity of $\lambda,\mu$
we thus get
$$
\lambda\left(M\times V\right)=\lim_{n}\lambda\left(\left[M\cap U_{n}\right]\times V\right)=\lim_{n}\mu\left(M\cap U_{n}\right)\nu\left(V\right)=\mu\left(M\right)\nu\left(V\right)
$$
for all Borel sets $M\subset X$ and all open sets $V\subset Y$ of
finite measure.
Now, define analogously
$$
\mathcal{H}_{m,M}:=\left\{ N\subset Y\,\mid\, N\text{ Borel and }\lambda\left(M\times\left[N\cap V_{m}\right]\right)=\mu\left(M\right)\cap\nu\left(N\cap V_{m}\right)\right\}
$$
for each Borel set $M\subset X$ of finite measure. Just as above,
we see that $\mathcal{H}_{m,M}$ contains all open sets and then all
Borel sets. Using the same limit argument as above, we get
$$
\mu\left(M\times N\right)=\mu\left(M\right)\nu\left(N\right)
$$
for all Borel sets $N\subset Y$ and all Borel sets $M\subset X$
of finite measure. A final limit argument (using $\sigma$-finiteness)
removes the assumption that $M$ has to have finite measure. $\square$