prove $\sup X \le \sup Y$ if $X$ is a subset of $Y$ both sets are nonempty and $Y$ is bounded above

Working through foundations of mathematical anlysis by johnsonbaugh per suggestion and wondering if the following proof works? (no solutions to book)

Problem:

Let X and Y be nonempty subsets of real numbers such that $X \subset Y$ and Y is bounded above. Prove that $\mathop{\mathrm{sup}} X \le \mathop{\mathrm{sup}} Y$.

Proof:

Given that Y is bounded above, by the least upper bound axiom there exists a least upper bound of Y. Given that $X \subset Y$ we have that $(\forall x \in X)(x \in Y)$. Then, since Y is bounded above, X must be and we have the existence of a least upper bound for X. Now, by definition of a supremum for every upper bound, $a \in Y$, $a\ \le \mathop{\mathrm{sup}} Y$. Hence, for every upper bound $b \in X$, we must have that $b \le \mathop{\mathrm{sup}} Y$ and it follows that $\mathop{\mathrm{sup}} X \le \mathop{\mathrm{sup}} Y$

The last bit is confused. You know that $\sup(Y)$ exists and is an upper bound for $X$ as well, as $X \subseteq Y$. So $\sup(X)$ exists. But $\sup(X)$ is the smallest of all upperbounds of $X$, and $\sup(Y)$ is one of the possible upper bounds. So..
• Not sure I follow $\mathop{\mathrm{sup}} Y$ is one of the possible upper bounds.. So, since $\mathop{\mathrm{sup}} X$ is the smallest of all upper bounds of X and $\mathop{\mathrm{sup}} Y$ is a possible upper bound it must be less than $\mathop{\mathrm{sup}} X$ by definition of $\mathop{\mathrm{sup}} X$ if im following? I thought that's what my last line stated? – Pythonista Feb 6 '16 at 14:25
• @Slayer $\sup Y$ is an upper bound for $Y$. Then $X\subseteq Y$ implies that it is an upper bound of $X$ as well. That implies directly that $\sup Y$ will not be smaller than the least upper bound of $X$. So $\sup X\leq \sup Y$. – drhab Feb 6 '16 at 14:40
• You said "Now, by definition of a supremum for every upper bound, $a \in Y$, $a \le \sup Y$", which makes no sense. – Henno Brandsma Feb 6 '16 at 14:42