The purpose of this answer is to recall that the classical proof of the existence of suprema of bounded families of reals (as found in any analysis textbook) continues to work without any modifications whatsoever for proper classes. (In particular, we do not assume that the existence of suprema of bounded nonempty subsets of reals has been proved yet.)

In fact, the proof works in the Zermelo set theory, without the axiom of replacement and the axiom of choice.

Theorem. Suppose $C$ is a nonempty class (possibly proper) and $g\colon C→{\bf R}$ is a map bounded from above. Then $\sup_{c∈C} g(c)$ exists.

Proof. Denote by $U$ the set of upper bounds for $g$, i.e., $U=\{u∈{\bf R}\mid \forall c∈C\colon f(c)≤u\}$, which exists by the axiom of separation.

Since $\sup_{c∈C} g(c)$ is by definition the smallest upper bound for $g$, we have to show that $U$ has the smallest element. By assumption, $U$ is nonempty. Since $C$ is nonempty, the set $U$ is bounded from below, namely, it is bounded from below by $g(c)$ for any $c∈C$. Thus, $U$ is a nonempty upward-closed subset of reals bounded from below, i.e., an upper Dedekind cut, so $\inf U$ exists. It remains to show that $\inf U ∈ U$.

Indeed, $\inf U ∈ U$ is equivalent to $$\forall c∈C\colon f(c) ≤ \inf U,$$ which in its turn (by definition of $\inf$) is equivalent to $$\forall c∈C \forall u∈U\colon f(c) ≤ u,$$ equivalently, $$\forall u∈U \forall c∈C\colon f(c) ≤ u,$$ i.e., $$\forall u∈U\colon u∈U,$$ which is tautologically true. Thus, $\inf U ∈ U$, so $\inf U$ is the smallest upper bound for $g$, which, therefore, exists.

The purpose of this answer is to recall that the classical proof of the existence of suprema of bounded families of reals (as found in any analysis textbook) continues to work without any modifications whatsoever for proper classes. (In particular, we do not assume that the existence of suprema of bounded nonempty subsets of reals has been proved yet.)

In fact, the proof works in the Zermelo set theory, without the axiom of replacement and the axiom of choice.

Theorem. Suppose $C$ is a nonempty class (possibly proper) and $g\colon C\to \mathbf R$ is a map bounded from above. Then $\sup_{c\in C} g(c)$ exists.

Proof. Denote by $U$ the set of upper bounds for $g$, i.e., $U=\{u\in\mathbf R\mid \forall c\in C\colon g(c)\le u\}$, which exists by the axiom of separation.

Since $\sup_{c\in C} g(c)$ is by definition the smallest upper bound for $g$, we have to show that $U$ has the smallest element. By assumption, $U$ is nonempty. Since $C$ is nonempty, the set $U$ is bounded from below, namely, it is bounded from below by $g(c)$ for any $c\in C$. Thus, $U$ is a nonempty upward-closed subset of reals bounded from below, i.e., an upper Dedekind cut, so $\inf U$ exists. It remains to show that $\inf U \in U$.

Indeed, $\inf U \in U$ is equivalent to $$\forall c\in C\colon g(c) \le\inf U,$$ which in its turn (by definition of $\inf$) is equivalent to $$\forall c\in C \forall u\in U\colon g(c) \le u,$$ equivalently, $$\forall u\in U \forall c\in C\colon g(c) \le u,$$ i.e., $$\forall u\in U\colon u\in U,$$ which is tautologically true. Thus, $\inf U \in U$, so $\inf U$ is the smallest upper bound for $g$, which, therefore, exists.

The purpose of this answer is to recall that the classical proof of the existence of suprema of bounded families of reals (as found in any analysis textbook) continues to work without any modifications whatsoever for proper classes.

In fact, the proof works in the Zermelo set theory, without the axiom of replacement and the axiom of choice.

Theorem. Suppose $C$ is a nonempty class (possibly proper) and $g\colon C→{\bf R}$ is a map bounded from above. Then $\sup_{c∈C} g(c)$ exists.

Proof. Denote by $U$ the set of upper bounds for $g$, i.e., $U=\{u∈{\bf R}\mid \forall c∈C\colon f(c)≤u\}$, which exists by the axiom of separation.

Since $\sup_{c∈C} g(c)$ is by definition the smallest upper bound for $g$, we have to show that $U$ has the smallest element. By assumption, $U$ is nonempty. Since $C$ is nonempty, the set $U$ is bounded from below, namely, it is bounded from below by $g(c)$ for any $c∈C$. It remains to show that $\inf U ∈ U$.

Indeed, $\inf U ∈ U$ is equivalent to $$\forall c∈C\colon f(c) ≤ \inf U,$$ which in its turn (by definition of $\inf$) is equivalent to $$\forall c∈C \forall u∈U\colon f(c) ≤ u,$$ equivalently, $$\forall u∈U \forall c∈C\colon f(c) ≤ u,$$ i.e., $$\forall u∈U\colon u∈U,$$ which is tautologically true. Thus, $\inf U ∈ U$, so $\inf U$ is the smallest upper bound for $g$, which, therefore, exists.

The purpose of this answer is to recall that the classical proof of the existence of suprema of bounded families of reals (as found in any analysis textbook) continues to work without any modifications whatsoever for proper classes. (In particular, we do not assume that the existence of suprema of bounded nonempty subsets of reals has been proved yet.)

In fact, the proof works in the Zermelo set theory, without the axiom of replacement and the axiom of choice.

Theorem. Suppose $C$ is a nonempty class (possibly proper) and $g\colon C→{\bf R}$ is a map bounded from above. Then $\sup_{c∈C} g(c)$ exists.

Proof. Denote by $U$ the set of upper bounds for $g$, i.e., $U=\{u∈{\bf R}\mid \forall c∈C\colon f(c)≤u\}$, which exists by the axiom of separation.

Since $\sup_{c∈C} g(c)$ is by definition the smallest upper bound for $g$, we have to show that $U$ has the smallest element. By assumption, $U$ is nonempty. Since $C$ is nonempty, the set $U$ is bounded from below, namely, it is bounded from below by $g(c)$ for any $c∈C$. Thus, $U$ is a nonempty upward-closed subset of reals bounded from below, i.e., an upper Dedekind cut, so $\inf U$ exists. It remains to show that $\inf U ∈ U$.

Indeed, $\inf U ∈ U$ is equivalent to $$\forall c∈C\colon f(c) ≤ \inf U,$$ which in its turn (by definition of $\inf$) is equivalent to $$\forall c∈C \forall u∈U\colon f(c) ≤ u,$$ equivalently, $$\forall u∈U \forall c∈C\colon f(c) ≤ u,$$ i.e., $$\forall u∈U\colon u∈U,$$ which is tautologically true. Thus, $\inf U ∈ U$, so $\inf U$ is the smallest upper bound for $g$, which, therefore, exists.