If $S$ is a nonempty set, is the following statement correct: $\varnothing \subset S$? It's confusing me becuase $\varnothing$ does not contain any elements so I'm struggling with the logic behind this statement.
Yes, it's correct. To say "$X \subseteq Y$" means that everything which is an element of $X$ is also an element of $Y$. Now imagine how you would demonstrate this wrong: you would supply an element of $X$ that is not an element of $Y$.
So, to show that $\emptyset \subseteq S$ is false, we would have to supply an element of $\emptyset$ that is not an element of $S$. Since we can't supply an element of $\emptyset$ at all, no such counterexample can be found; so the statement is true.
Now, the symbol $\subset$ denotes proper subset - $X \subset Y$ if and only if $X \subseteq Y$ and $X \neq Y$. So $\emptyset \subset S$ if and only if $\emptyset \subseteq S$ (which is always true) and $\emptyset \neq S$ (which is true if $S$ is nonempty).
The statement is vacuously true: suppose the proposition is false: therefore $\emptyset \not\subset S$. Therefore, there must exist $s\in \emptyset$ such that $s \not\in S$. But $\emptyset$ does not contain any element, meaning $s$ cannot exist. It must be true because stating otherwise is impossible.
$\varnothing\subset S$ for any set $S$, since you cannot find a counter-example, even if $S=\varnothing$: an element in $\varnothing$ which would not be in $S$.
Note: The correct mathematical notation is $\varnothing$ (code
\varnothing), not $\emptyset$.
It is true because it is not false that every object in $\emptyset$ belongs to $S$.
Let me elaborate.
$x \in \emptyset$ but $\emptyset$ is empty so whatever you want to say about any object inside $\emptyset$ will always be true since none exist, you might as well say that $x \in \emptyset$.