Is my proof for $\sup(A) \le \sup(B)$ if $A \subseteq B$ correct?

Question

Suppose $A$ and B are sets such that $A \subseteq B$. Now suppose $\sup(A)$ and $\sup(B)$ exist and $\sup(A) = M$ and $\sup(B) = N$. Therefore, we have that $\forall \epsilon > 0 \exists x \in A \space M - \epsilon < x < M$ and $\forall \epsilon > 0 \exists y \in B \space N - \epsilon < y < N$. For the sake of contradiction, suppose $\sup(A) > \sup(B)$. Since $M > N$ and $M - \epsilon > N - \epsilon$ it follows that $\exists x \in A \space \forall y \in B \space x > y$. However, since $A \subseteq B$ we can find a $y \in B$ such that $y = x$, which contradicts the fact that $\exists x \in A \space \forall y \in B \space x > y$. Therefore, $\sup(A) \le \sup(B)$.

Answer 1

The conditions on $M$ are that

1. for all $x\in A$, $x\le M$
2. for all $\varepsilon>0$, there exists $x\in A$ such that $M-\varepsilon<x$

Note the $\le$ sign and the difference in quantifiers. You cannot say in general that from $x\in A$ it follows that $x<M$: think to $M=\{0\}$, whose supremum is obviously $0$.

Your idea to go by contradiction is good (but you could avoid it). However $M-\varepsilon>N-\varepsilon$ does not imply so directly that there is $x\in A$ so that $x>y$ for every $y\in B$. For one thing: what's $\varepsilon$? You have to pick one, precisely $\varepsilon=M-N$. Here's how you can fix the rest of the proof.

From $M>N$ it follows that $\varepsilon=M-N>0$ and therefore there exists $x\in A$ with $$ x>M-\varepsilon=M-(M-N)=N $$ from property 2 applied to $A$; from property 1 applied to $B$, we know that $y\le N$, for every $y\in B$. Therefore, for every $y\in B$ we have $y\le N<x$ and so $y<x$. Thus $x\notin B$: a contradiction.

Note 1. Saying “since $A\subseteq B$ we can find a $y\in B$ such that $y=x$” is a very convoluted form of saying that $x\in B$.

Note 2. If the supremum of $B$ is $\infty$, then of course $\sup(A)\le\sup(B)$, so you can assume $\sup(B)$ is finite. If $\sup(A)=\infty$, then it cannot be that $\sup(B)$ is finite, because for every $K$ there exists $x\in A$ (hence $x\in B$) with $x>K$.

Answer 2

In my opinion, there are 2 issues with your proof: it relies on the finiteness of the suprema and it's unnecessarily complicated. You can work directly from definition: $$ \sup B\geq x,\forall x\in B\implies \sup B\geq x,\forall x\in A\implies\sup B\geq \sup A. $$ The first implication uses $A\subset B$ and second uses the definition of supremum.