I found the following result at the beginning of some notes on topological vector spaces (TVS). This is a quite fundamental result, that apparently is considered the corresponding version of the triangle inequality in TVS. Here there is the result with two proofs.
Theorem: Let $X$ be a TVS. Then, for every $W \in \mathcal{N}_0$ there exists a $U \in \mathcal{N}_{0}^\text{sym}$ such that $U+U \subset W$.
Proof 1: For any neighborhood $W \in \mathcal{N}_o$, there are neighborhoods $U_1, U_2$ of $0$ such that $U_1 + U_2 \subseteq W$. The result follows by setting $U = U_1 \cap U_2 \cap (-U_1) \cap (-U_2)$.
Proof 2: Since $0+0 = 0$ and addition is continuous, there exist neighborhoods $U_1, U_2 \in \mathcal{N}_0$ such that $U_1 + U_2 \subseteq W$. The result follows by setting $U = U_1 \cap U_2 \cap (-U_1) \cap (-U_2)$, by noting that $U$ is symmetric.
Here $\mathcal{N}_{0}$ denotes the neighborhood of $0$, and $\mathcal{N}_{0}^\text{sym}$ the symmetric neighborhood of $0$.
Now, in both cases I am not sure about the logic behind them.
In particular, here there are my doubts:
1) In proof 1 how do we come up with the fact that there are neighborhoods $U_1, U_2$ of $0$ – and not of something else! – such that $U_1 + U_2 \subseteq W$?
2) In proof 2 why do we focus on $0 + 0=0$? Is it because if we would have focused on – let's say– $2 + (-2) = 0$, we would have got still a neighborhood of $0$ by translation invariance, i.e. $\mathcal{N}_0 +2$?
As you could notice, both questions are fairly similar and are concerned, in my opinion, with the inner logic of TVS, that completely baffles me right now.
Thus, any feedback or help will be most welcome.
Thank you for your time.
1) Because the addition is continuous at $(0,0)$ (being continuous), you can find a neighbourhood $W'$ such that $W'+W'\subseteq W$, and then, just take $W\cap W'$.
2)This exactly because for any neighbourhood $W$ of $x$, $-x+W$ is a beighbourhood of $0$.