Continuity of $F: X \times Y \to Z$ in terms of nets Let $F: X \times Y \to Z$ be a map where $X,Y,Z$ are topological spaces and the domain has the product topology.
Prove that the following statements are equivalent:
(1) $F$ is continuous
(2) For every couple of nets $(x_\alpha)_{\alpha \in I}$ in $X$ and $(y_\beta)_{\beta \in J}$ in $Y$ with $x_\alpha \to x$ and $y_\beta \to y$, we have $F(x_\alpha, y_\beta) \to F(x,y).$
Here $((x_\alpha, y_\beta))_{(\alpha, \beta) \in I \times J}$ is the net with partial order on $I \times J$ defined by 
$$(\alpha, \beta) \leq (\gamma, \delta) \iff \alpha \leq \gamma, \beta \leq \delta$$
I managed to show that $(1) \implies (2)$. 
I'm currently trying to prove that $(2) \implies (1)$. I know that a function is continuous iff all nets converging in the domain have image nets converging in the codomain with limit the image of the limit in the domain, so I guess I must use this fact.
So, let $((x_\alpha, y_\alpha))_{\alpha \in K}$ be a net in $X \times Y$ converging to $(x,y)$. By product topology, we then know that the nets $(x_\alpha)_{\alpha \in K} \to x$ and $(y_\alpha)_{\alpha \in K} \to y$ and then it follows that 
$$(F(x_\alpha, y_\beta))_{\alpha, \beta \in K} \to F(x,y)$$
Am I even on the right track? How can I finish this proof?
 A: $(2)\implies(1)$,
$(\alpha,\beta)\le(\gamma,\delta)\implies (x_\gamma,y_\delta)\in U \text{ (U neighborhood of $(x,y)$) }\implies F(x_\gamma,y_\delta)\in V \text{ (V neighborhood of $F(x,y)$) }$
Let $K=X\times Y$ and $k=(x,y)\in K$, $k_U=(x_\alpha,y_\beta)$ be the convergent net.
Suppose that $(1)$ fails i.e. $F$ is not continuous, then there exists one open neighborhood $V\subset Z$ of $F(x,y)-F(k)$ s.t. $F^{-1}(V)$ is not open in $K$. Therefore $\exists!U\subset K$ where $U$ is an open neighborhood s.t. $U$ is an open neighborhood of $(x,y)=k$ contained in the inverse image of $V$, which means for each $U$, $\exists (x_a,y_b)\in U\subset K\implies F(x_a,y_b)\notin V\implies F(x_\alpha,y_\beta)\neq F(x,y)=F(k)$
Now, construct a directed set and a net to see if our assumption fails.
Let $J=\{U|\text{ $U$ open neighborhodd of $(x,y)$ }\}$ and let it be ordered by "$\supset$". Thus the net $(k_U)_{U\in J}$ is a net converging to $k$, so $U\subset U_o\implies k_U\in U\subseteq U_o$ by the definition of convergent nets.
But, F(k_U) doesn't converge to $F(k)$ since we have argued in the first paragraph that $F(k_U)\notin V$, contradicts to our assumption that two nets are both convergent. (I start using $k$ because the initial expression contains so many characters...)

I tried to minimize the proof above but I'm not very sure if this is ambiguous.
Choose an open set $V\subset Z$, we wish to prove $F^{-1}(V)$ open in $X\times Y$. If $V=\varnothing$ then it's obviously true. So let's assume $F(x,y)\in V$, then there exists some open subset of $V$ s.t. $B\subseteq V$, therefore, $V$ is a neighborhood of $F(x,y)$. Here, $B$ is determined by the relation $(\alpha,\beta)\le(\gamma,\delta)$ to include some other point $F(x_\gamma,y_\delta)$ in the net because the net is convergent. By the assumption, an open set $A$ contains both $(x,y)$ and $(x_\gamma,y_\delta)$ ($(\alpha,\beta)\le(\gamma,\delta)$) and $A\subseteq F^{-1}(V)$. If not, then $(x_\gamma,y_\delta)$ may be outside of the neighborhood and contradicts our assumption. Since this hold for every point in $F^{-1}(V)$, every point of $F^{-1}(V)$ is its interior point, so $F^{-1}(V) $ is open. 
A: If you already know the characterisation of the product topology using nets (net convergence in the product is equivalent to the projection nets converging to the projected limit) and the general continuity criterion by nets, then (2) implies (1) is quite straightforward, and you were almost there. I'll just write nets as functions (what they are by definition) and in "sequence-like" index-notation (as you did) to hopefully make my point clearer:
Let $n: I \to X \times Y$ be a net converging to $(x,y) \in X \times Y$. (This can be denoted by $(x_i, y_i)_{i \in I} \to (x,y)$.)
Then $\pi_X \circ n: I \to X$ converges to $x$ (or $(x_i)_{i \in I} \to x$) and $\pi_Y \circ n: I \to Y$ converges to $y$ (or $(y_i)_{i \in I} \to y$.)
Now we apply (2): for the product directed set $I \times I$ (in the product order, as defined), the net $f \circ ((\pi_X \circ n) \times (\pi_Y \circ n)):I \times I \to Z$ converges to $f(x,y)$ (also written as $(f(x_i, y_j))_{(i,j) \in I \times J} \to f(x,y)$. 
We can define $\Delta: I \to I \times I$ by $\Delta(i)=(i,i)$ and it's easy to see that $\Delta$ is monotonous and has a cofinal image in $I \times I$ (because for $(i,j) \in I \times I$ we can find $k \in I$ such that $k \ge i, k \ge j$ and then $\Delta(k) = (k,k) \ge (i,j)$, so this follows simply by directedness of $I$) and so $$(f \circ ((\pi_X \circ n) \times (\pi_Y \circ n))) \circ \Delta = f \circ n$$ from $I$ to $Z$ is a subnet of the previous net and so also converges to $f(x,y)$, or in index-terms $(f(x_i, y_i))_i \to f(x,y)$, which is as required to show net-continuity and (hence!) continuity.
Of course the real work of this fact is in the proposition that continuity of a function is equivalent to net-continuity. The rest is just "index juggling", in essence.
