Historically, "function" meant something like an element of the vector space $C(\mathbb{R})$ of continuous functions $\mathbb{R} \to \mathbb{R}$, "functional" meant something like a linear functional $C(\mathbb{R}) \to \mathbb{R}$ (that is, a thing which takes functions as input and returns numbers), and "operator" meant something like a linear transformation $C(\mathbb{R}) \to C(\mathbb{R})$ (that is, a thing which takes functions as input and returns functions).
Of course, from the modern point of view there's no real reason not to call operators functions, since we recognize sets of functions as after all just sets so we can talk about functions in and out of them. It's just convenient in functional analysis to invoke a certain context by using the word "operator."
You're more or less correct about what an operation is.
Top is the category of topological spaces and continuous maps simply by definition; topology typically deals with continuous maps, making this category the most important one, and thus by convention it's the one meant when saying "the category of topological spaces".
(aside: other conventions on what Top or "the category of topological spaces" stands for are far more likely to disagree on what the objects are, rather than the morphisms. e.g. to make the objects be merely the compactly generated Hausdorff spaces)
You can, of course, make all sorts of other categories. The category of topological spaces and open maps is a perfectly reasonable category to make; it's just less useful.
It takes a bit to get used to, but category theory rejects the mindset that mathematics is about objects, with the mappings between objects being a derived notion. Instead, you need to consider objects and mappings as equals -- or even to consider the objects superfluous.
On that last point, my favorite example of a category whose emphasis is on the morphisms is matrix algebra. The set of all matrices, with composition defined by multiplication, form a category. (with addition, you get an Abelian category) The objects of this category really play no role beyond bookkeeping to say which matrix products are defined.
(This category is, of course, equivalent to the category of fintie-dimensional vector spaces and linear maps)
Best Answer
The answer to the first question is clearly no, since the mapping can collapse the domain to the zero vector. A function that simply interchanges two points has an inverse that takes bounded sets to bounded sets, but the function is neither continuous nor linear.
The second statement is precisely equivalent to the first, so it does indeed mean that the image of a bounded set under a continuous linear mapping is bounded. The answer to the final question is no, just as in the first part.