I have a question regarding the property of topological vector spaces $X$ that their topology is completely determined by the neighbourhood system of any point $x\in X$ (and therefore in particular by the neighbourhood system of the origin.
Translations in a tvs are continuous which means that it maps open sets to open sets.
1.) Does that already imply that when I have an open set $U\subseteq X$ and translate that by an element $x\in X$, that $U+x$ is also open? I am not certain because the translation is defined pointwise: $T_{x_0}(x)= x+x_0$. So can I conclude that if I do that pointwise translation for any element in $U$ that the set $\{x+x_0|x\in U\}$ is again an open set? Or do I have to prove anything here?
2.) How does the fact that open sets are translated to open sets tell me that neighbourhoods are mapped to neighbourhoods? A neighbourhood is a set $V\subseteq X$ that contains an open set $U\subseteq X$. But how do I know what happens with all the elements of the neighbourhood $V$ that do not belong to the open set $U$ that is contained in $V$?
3.) How can I now show that the neighbourhood system of the origin somehow is a basis for the topology? I.e., how does a neighbourhood basis relate to a basis or subbasis of the topology?
No, continuous functions don't have to map open sets to open sets. The fact that it works for translations is because translations are homeomorphisms, which is a lot stronger condition than just continuous. And indeed, homeomorphisms map open subsets to open subsets.
A translation $x\mapsto x+x_0$ is a homeomorphism because it has a (continuous) inverse $x\mapsto x-x_0$.
And the other important property of translations is that for any $x,y\in X$ there is a translation that maps $x$ to $y$. And this pretty much means that topology around any point is the same.
More generally, spaces with the property that for any $x,y\in X$ there is a homeomorphism mapping $x$ to $y$ are called homogeneous spaces and indeed they have this nice property that they look the same wherever you are. Thus what we see here is that topological vector spaces are homogenous.
Its like a weird planet, wherever you land everything looks the same. To the point that you can't really tell where you are. Sounds bad, but for mathematicians it is good, because the space behaves better in some sense.
Everything in maths has to be proven. Yes, it is correct. But not because translation is continuous, but because it is a homeomorphism.
And what is supposed to happen to them? They are irrelevant. They are mapped somewhere, but they don't matter much for the definition of neighbourhood.
Also you should pay attention to the definition of "neighbourhood". Because some authors by "neighbourhood" mean "open neighbourhood", which is easier to work with. In particular when we say that neighbourhoods form a topology then we mean open neighbourhoods.
It is not. The statement "the neighbourhood system of the origin determines the basis" means that any open subset arises somehow from neighbourhoods of the origin. And this "somehow" in this case means "by translation". So for example, if $(-11, -9)$ is an open subset of $\mathbb{R}$ then we can write it as $(-1,1)-10$ and $(-1,1)$ is a neighbourhood of the origin $0$. And so the knowledge of neighbourhoods of $0$ is enough to determine any open subset of $\mathbb{R}$, even though most of them obviously don't contain $0$.
Note that this is not necessarily true for other spaces. For example consider the closed interval $[0,1]$. Neighbourhoods of $0$ don't determine neighbourhoods of say $1/2$ and vice versa. This space is not homogenous, and indeed, every homeomorphism $[0,1]\to [0,1]$ has to map $\{0,1\}$ to $\{0,1\}$. Meaning $0$ and $1$ differ substantially from other points. Which is intuitively clear: these are ends of $[0,1]$ after all, so they have to be special. But now you have this intuition formalized.