Is $1$ a subset of $\{1\}$
$1$ is an element of $\{1\}$, $\{1\}$ is a subset of $\{1\}$, $\{1\}$ is an element of $\{\{1\}\}$ and $\{\{1\}\}$ is a subset of $\{\{1\}\}$.
$1$ is generally not a subset of $\{1\}$, since $1$ is a natural number (or a real number, or whatever) and not a set. These objects are of two different types.
But there is something to be said here. We can represent numbers using sets. We can declare that $0$ is $\varnothing$, and that $1=\{0\}$ or $\{\varnothing\}$, and that $2=\{0,1\}$ and so on. Then a number is a set.
Still that doesn't mean that $1\subseteq\{1\}$. This would very much depend on the representation of $1$ as a set.
So while "working mathematics" is typed (meaning the type of objects matters), we can also work in an untyped environment, where everything has the same type (for example, everything is a set).
As @AsafKaragila already said, you can define natural numbers as sets. In fact, in most axiomatic set theories, that's the only way of defining numbers, since every element of a set is a set.
As for your specific question, suppose that $1\subset\{1\}$. This leaves us with two options:
$1=\emptyset$ and the inclusion trivially holds.
$1\neq\emptyset$. Since $\{1\}$ has only two subsets, $1=\{1\}$.
Option 1 is possible, since identifying $1$ with the empty set is perfectly valid, but it's much more natural to identify $0$ with the empty set.
Option 2 just "looks wrong", but in elementary set theory, we can't really make any statements beyond that. In any axiomatic set theory that includes the Axiom of Regularity, $1=\{1\}$ is false since $x\notin x$ for every set $x$.