Is [7, 10) a subset of {7, 8, 9, 10}? I'm learning elementary set theory and so, in an effort to understand my reasoning, I'd like some feedback. I apologise if it's too basic - I'm barely starting out and don't want to get my foundation wrong. 
I'm asked the following: 

Is it true that [7, 10) ⊆ {7, 8, 9, 10}?

I've dubbed {7, 8 , 9, 10} = A for simplicity's sake. The interval [7, 10) is composed of the numbers 7 ≤ x < 10, which include 7.1, 7.11, 7.222, and an infinite amount of numbers before 10. Meanwhile, A is composed of only four elements: 7,8,9,10. Simultaneously, elements of A are also present in the interval: 7, 8, 9 are, indeed, part of [7, 10) and of A. Would this be enough to consider [7, 10) ⊆ {7, 8, 9, 10} as true, or does the presence of other numbers (like those I listed above) disqualify the interval from being a subset of A? 
 A: $A = [7, 10)$
$B = \{7, 8, 9, 10 \}$
$A$ is not a subset of $B$ because $7.1 \in A, 7.1 \notin B $ , 
$B$ is a not a subset of $A$ because $10 \in B, 10 \notin A$ 
But $B$ is a subset of $[7, 10]$
A: The set $[7,10)$ is an interval and contains an infinite number of elements: all real numbers from 7 included to 10 excluded. That makes an infinity of numbers. Set theorists say that there are more real numbers in any  one unit long interval than in the whole ( infinite) set of natural numbers. 
See: § " Sets with the cardinality of the continuum". https://en.wikipedia.org/wiki/Cardinality_of_the_continuum
The set $\{7,8,9,10\}$ only has 4 elements, and is therefore a finite set. 
An infinite set cannot be a subset of a finite set. 
As a finite set the set $\{7,8,9,10\}$ could be a subset of $[7,10)$ , but it's not the case, since it is not true that all the elements of $\{7,8,9,10\}$ are also elements of $[7,10)$. 
True, the numbers 7, 8, 9 are also elements of $[7,10)$, but number 10 is not. 
A: Let's call your interval $I$.
Intuitively, you can think of sets as bags that don't contain duplicated items.
If $A$ out-contain every items in the bag $I$ then we say $I \subset A$.
In your example, you have listed some elements that existed in the $I$ but not in the set $A$. And thus, $I \not\subset A$.
Since you said you're learning elementary set theory, it's a good habit to ask yourself whether $I \subset A$ or not.
And if you have learned about the intersection operation $\cap$ then you could try to ask what is $I \cap A$. And then try to do the same with examples where there are subsets relation, maybe equal set relation. You'll find it make much more sense :)
A: Absolutely and definitely no!
The set $A=\{7, 8, 9, 10\}$ is a subset of $\mathbb{N}$ or $\mathbb{Z}$ and has only four elements, this is the four integers $7$, $8$, $9$ and $10$.
The set $B=[7, 10)$ is meant to be a real interval, it is the set of all the real numbers $x$ such that $7\le x<10$.
The cardinality of $A=\{7, 8, 9, 10\}$ is $4$, while the cardinality of $B=[7,10)$ is the same of $\mathbb{R}$, which is infinite and it is even a bigger infinite (cardinality of the continuum) than the cardinality of $\mathbb{N}$.
Notice that $10$ is an element of $A$ but not of $B$, while any irrational number between $7$ and $10$ belongs to $B$ but not to $A$.
If $B$ was a subset of $A$, this would mean that $B\cap A=B$. But in this case we have that
$$
B\cap A=\{7,8,9\}
$$
is a non-trivial subset of both $A$ and $B$.
I think this question rised from an uncomplete understanding of the notation: $[a,b]$, $[a,b)$, $(a,b]$ and $(a,b)$ denote connected subsets of $\mathbb{R}$, while $\{a,\ldots,b\}$ an unconnected one.
