Equivalence of Definitions of Hereditarily Compact

Theorem
Let $(X,\tau)$ be a topological space.

Then $(X,\tau)$ is hereditarily compact if and only if for each family $(U_i)_{i\in I}$ of open sets, there exists a finite subset $J\subset I$ such that:

$$ \bigcup_{j\in J} U_j = \bigcup_{i\in I} U_i. $$

Proof
Necessary condition

Let $Y\subset X$ be a subspace of $(X,\tau)$.

Let $(V_i)_{i\in I}$ be an open cover for $Y$, i.e.

$$ \bigcup_{i\in I} V_i = Y. $$

Then by definition of the subspace topology, we find

$ V_i = U_i\cap Y$ for a certain $V_i \in \tau$.

But then $(U_i)_{i\in I}$ is a family of open sets, and by hypothesis, there exists a finite subset $J\subset I$ such that:

$$ \bigcup_{j\in J} U_j = \bigcup_{i\in I} U_i. $$

But then we have

$$ \bigcup_{i\in I} V_i = \bigcup_{i\in I} (U_i \cap Y) = (\bigcup_{i\in I} U_i) \cap Y = (\bigcup_{j\in J} U_j) \cap Y = \bigcup_{j\in J} (U_j \cap Y) = \bigcup_{j\in J} V_j. $$

Thus $(V_j)_{j\in j}$ is a finite subcover of $(V_i)_{i\in I}$ for $Y$.

Thus $Y$ is a compact subspace of $(X,\tau)$.

Sufficient condition

Let $(U_i)_{i\in I}$ be any collection of open sets in $(X,\tau)$.

Clearly, we see that:

$$\bigcup_{i\in I} U_i \subset X. $$

By hypothesis, then $\bigcup_{i\in I} U_i$ is compact as a subspace of $(X,\tau)$.

Furthermore, $U_i = U_i \cap \bigcup_{i\in I} U_i$ is open in $\bigcup_{i\in I} U_i$, by definition of the subspace topology.

Therefore, $(U_i)_{i\in I}$ is an open cover for $\bigcup_{i\in I} U_i$.

Since $\bigcup_{i\in I} U_i$ is compact, there exists a finite subcover of $(U_i)_{i\in I}$ for $\bigcup_{i\in I} U_i$, say $(U_j)_{j\in J}$.

Then by definition of open cover, $\bigcup_{j\in J} = \bigcup_{i\in I} U_i$.