$\sup_{i \in \mathbb{N}} \sup_{j \in \mathbb{N}} B_{ij} = \sup_{j \in \mathbb{N}} \sup_{i \in \mathbb{N}} B_{ij}$ 
For any double sequence $B_{ij} \ i,j \in \mathbb{N}$ of real numbers we have $$\sup_{i \in \mathbb{N}} \sup_{j \in \mathbb{N}} B_{ij} = \sup_{j \in \mathbb{N}} \sup_{i \in \mathbb{N}} B_{ij}$$

Here is my attempt.
$\forall m,n \in \mathbb{N}$ it holds that
$$
B_{m,n} \leq \sup_{j \in \mathbb{N}} \sup_{i \in \mathbb{N}} B_{i,j}.
$$
The right hand side is independent of $n$ thus it holds that
$$
\sup_{n \in \mathbb{N}} B_{m,n} \leq \sup_{j \in \mathbb{N}} \sup_{i \in \mathbb{N}} B_{i,j}.
$$
The right hand side is independent of $m$ thus it holds that 
$$
\sup_{m \in \mathbb{N}} \sup_{n \in \mathbb{N}} B_{m,n} \leq \sup_{j \in \mathbb{N}} \sup_{i \in \mathbb{N}} B_{i,j}.
$$
Then for $m=i, \ n=j$ we get that
$$
\sup_{i \in \mathbb{N}} \sup_{j \in \mathbb{N}} B_{i,j} \leq \sup_{j \in \mathbb{N}} \sup_{i \in \mathbb{N}} B_{i,j}.
$$
Now similiar $\forall m,n \in \mathbb{N}$ it holds that 
$$
B_{m,n} \leq \sup_{i \in \mathbb{N}} \sup_{j \in \mathbb{N}} B_{i,j}.
$$
The right hand side is independent of $m$ thus it holds that
$$
\sup_{m \in \mathbb{N}} B_{m,n} \leq \sup_{i \in \mathbb{N}} \sup_{j \in \mathbb{N}} B_{i,j}.
$$
The right hand side is independent of $n$ thus it holds that 
$$
\sup_{n \in \mathbb{N}} \sup_{m \in \mathbb{N}} B_{m,n} \leq \sup_{i \in \mathbb{N}} \sup_{j \in \mathbb{N}} B_{i,j}.
$$
Then for $m=i, \ n = j$
$$
\sup_{j \in \mathbb{N}} \sup_{i \in \mathbb{N}} B_{i,j} \leq \sup_{i \in \mathbb{N}} \sup_{j \in \mathbb{N}} B_{i,j}.
$$
Thus they are equal.
 A: As I posted in my comment, your proof is correct. This is how I would do it by first proving a more general lemma as follows. I admit that your proof is still easier (and less complicated)!
We will first prove that, given a family of sets $\left ( S_i \right )_{i=1}^{\infty}$,
$$\sup \bigcup_{i=1}^{\infty} S_i = \sup \left \{ \sup S_i \right \}_{i=1}^{\infty}.$$
That is indeed true, for $\sup \left \{ \sup S_i \right \}_{i=1}^{\infty}$ is an upper bound of every $S_k$ ($k=1,2, \dots$) and thus an upper bound of $\bigcup_{i=1}^{\infty} S_i$, therefore
$$\sup \bigcup_{i=1}^{\infty} S_i \leq \sup \left \{ \sup S_i \right \}_{i=1}^{\infty};$$
the converse also holds since $S_k \subseteq \bigcup_{i=1}^{\infty} S_i$ for each index $k$ and so $\sup \bigcup_{i=1}^{\infty} S_i$ is an upper bound of $S_k$, yielding
$$\sup \bigcup_{i=1}^{\infty} S_i \geq \sup \left \{ \sup S_i \right \}_{i=1}^{\infty}$$
as claimed.
After all the above, if $(b_{ij})$ is a double series (with $i,j \in \mathbb{N} = \{1,2, \cdots \}$), let $S_i$ be the set of all the terms of $(b_{ij})_{j=1}^{\infty}$. Then
$$\sup \bigcup_{i=1}^{\infty} S_i = \sup \left \{ \sup S_i \right \}_{i=1}^{\infty},$$
which means that
$$\underset{i,j \in \mathbb{N}}{\sup} b_{ij} = \underset{i \in \mathbb{N}}{\sup} \underset{j \in \mathbb{N}}{\sup} b_{ij},$$
as we wished. In a similar manner we derive that
$$\underset{i,j \in \mathbb{N}}{\sup} b_{ij} = \underset{j \in \mathbb{N}}{\sup} \underset{i \in \mathbb{N}}{\sup} b_{ij}$$
too, and the proof is complete.
Also, I consider it silly when you have to write essentialy the same proof more than once in order to change just some small details (like showing that what you did for index $j$ can also be done for index $i$), so you might just say that the second part of the proof is similar to the first. However, note that some professors strictly insist on writing full proofs, so ask them first before you try that!
