Proving by contradiction that $A \subset B \implies \sup(A) \leq \sup(B)$, for $A$ and $B$ bounded subsets of $\Bbb{R}$ Yesterday I saw a question related to calculus 1 in which they asked for a proof for the following:

Let $A$ and $B$ be two bounded subsets of $\mathbb R$.
$$A \subset B \implies \sup(A) \leq \sup(B)$$

The moment I saw it, reasoning by reductio ad absurdum came to my mind. So my personal proof would be:

Let's assume that the assertion
$$A \subset B \implies \sup(A) > \sup(B)$$
is true and we find a contradiction: $\sup(A) > \sup(B)$, since $\sup(B)$ is the smallest upper bound of $B$ and $\sup(A)$ strictly greater than $\sup(B)$ then this means that $\sup(A)$ which can belong to set $A$ does not belong to set $B$ and this is a contradiction because $A$ is included in $B$.
So we have shown by contradiction that the assertion $A \subset B \implies \sup(A) \leq \sup(B)$ is true.

Many teachers proved the same assertion with direct reasoning, but I preferred the absurd reasoning. What do you think about it? Is my proof true?
Thanks already.
 A: The proof isn't correct.  There's a problem with the word "can" in the phrase "sup $A$ which can belong to $A$."   "Can" is not enough.  Take a simpler example:
Prove that if $x < y$ then $x^2 < y^2$.
First note that the statement is not true.  We have $-3 < -2$ but $(-3)^2 > (-2)^2$.   But pretend we didn't notice that and we use your method of proof:
Since $x$ and $y$ can both be positive, we have $x^2 < xy < y^2$, therefore we're done.
Nope.  The statement is true in special cases, but not in all cases.  "Can" takes to special cases.   I guess your prove works in the special case that sup $A$ does belong to $A$, but what about the other cases?
A: As others have pointed out, your argument contains some flaws. That being said, here's how I would do a (very verbose) contradiction argument: assume that $A \subset B$, where $B$ is a bounded subset of $\mathbb{R}$. Assume, to the contrary, that $\sup A > \sup B$. Because $A \subset B$, it follows that $a \in A \Rightarrow a \in B$. By definition, for every $$b \in B, b \le \sup B < \sup A,$$ by our assumption. However, this also implies that $\sup B$ is an upper bound of $A$. That is, for every $a \in A, a \le \sup B < \sup A$, and so $\sup B$ is a smaller upper bound than $\sup A$, which is impossible!
