Ok, for this problem we have to prove that if a function is continuous, it is continuous in each variable separately.
What this means is that for some continuous function defined as:
$$f: \ X \ \times \ Y \ \rightarrow \ Z$$
This function is continuous in each variable if for each $y_0 \ \in \ Y$, we have $h: X \ \rightarrow \ Z$ defined as
$h(x) \ = \ F(x \ \times \ y_0)$ that is continuous. In addition, for each $x_0 \ \in \ X$, we have a function $k: Y \ \rightarrow \ Z$, defined as
$k(y) \ = \ F(x_0 \ \times \ y)$.
Let's look at the function $h$ first. Two things are happening here. First, we take some general element of $X$, call it $x$. Our function $h$ is first mapping $x$ to
$x \ \times \ y_0$, and then this is getting mapped to some element of $Z$. We can write this as two separate functions, called $h_1$ and $h_2$. We have:
$$h_1 : X \ \rightarrow \ X \ \times \ y_0$$
$$h_2: X \ \times \ y_0 \ \rightarrow \ Z$$
Where $h_2$ is the same as $F$. For the first function, this is the conventional map to products. As you may recall, we proved in a previous theorem that for some function of the form $f : A \ \rightarrow \ X \ \times \ Y$,
then the two "components" of $f$ must be continuous, meaning $f_1 : A \ \rightarrow \ X$ is continuous and $f_2 : A \ \rightarrow \ Y$ is continuous. For this function, we have:
$$h_{11} : X \ \rightarrow \ X$$
$$h_{12} : X \ \rightarrow \ y_0$$
Both of these function are continuous, the first is the inclusion function from $X$ to itself. The second one is the constant function from $X$ to $y_0$. Both functions of
this form are always continuous. Let's look at $h_2$ now. $X \ \times \ y_0$ is simply a domain restriction of the original domain of $F$, which is $X \ \times \ Y$.
Now, we assert that a domain restriction of a continuous function will still yield a continuous function. Let us have some general continuous function:
$$g : A \ \rightarrow \ B$$
We then restrict the
domain to some subset of $A$, called $S$. Our function is now $g : S \ \rightarrow \ B$. We can map the subset $S$ to $A$, as this is just the inclusion mapping, which is known
to be continuous. Our function with the domain restriction can then be written as:
$$i : S \ \rightarrow \ A \ \Rightarrow \ g : A \ \rightarrow \ B$$
This is the same thing as $(g \ \circ \ i)$.
We know that both $i$ and $g$ are continuous, so then their composition is as well, making our domain-restircted function also continuous. We know that $F$ is continuous, so from this, we know that
$h_2$ is also continuous. We now can say that our original function $h$ can be constructed in terms of continuous function $h_1$ and $h_2$ as:
$$h \ = \ (h_2 \ \circ \ h_1)$$
Which is a composition of continuous functions, therefore $h$ is continuous, and the function is continuous in each variable separately. The exact same process applies for proving
this in the case of $k$.