Simple question about Open Mapping Theorem's proof in Functional Analysis

Regarding the OMT's proof, I have a really simple question, but I can't found an acceptable answer for it, every source simply ommits it (for this motive I think it's really simple).
So, let's state the theorem: Let $E$ and $F$ be Banach spaces and $T \in \mathscr{L}(E,\,F)$, in which $\mathscr{L}(E,\,F)$ denotes the space of continuous linear operators from $E$ to $F$. If $T$ is surjective, then $T$ is an open mapping.
Alright, so the proof begins like this, we cover $E$ with the open balls of the form $B(0,\,n)$, so that $E = \bigcup_{n\,=\,1}^\infty B(0,\,n)$.
Since $T$ is onto, we have $T(E) = F = \bigcup_{n\,=\,1}^\infty T(B(0,\,n))$. Now is the part that I didn't quite understand, why we can "take" the closure of the aforementioned union?
In the proof I'm reading, the author simply takes the closure, getting $F = \bigcup_{n\,=\,1}^\infty \overline{T(B(0,\,n))}$, without further explanations.
I know that, since $\left\{T(B(0,\,n))\right\}_{n\,=\,1}^\infty$ covers $F$, "bigger" sets like $\overline{T(B(0,\,n))}$ would also cover it, but I was not able to write this observations as a proper proof.
Like I said, I think it's really simple, but it's bugging me and I don't want to simply go over this fact.
Any help is appreciated!

Solution 1:

All images of $T$ lie in $F$ by definition; we also take the closure in $F$ of that image. So if you want to be formal (using $A \subseteq \overline{A}$ for all unionants):

$$F\subseteq \bigcup_{n=1} T(B(0,n)) \subseteq \bigcup_{n=1} \overline{T(B(0,n))} \subseteq F$$

so we have a double inclusion and so $\bigcup_{n=1} \overline{T(B(0,n))}=F$.