Let $\{V_\alpha\}_{\alpha \in A}$ be an open cover of $C$. For each $\alpha$, $V_\alpha$ is open in the subspace topology on $C$, so $V_\alpha = U_\alpha \cap C$ for some open set $U_\alpha$ in $X$.

Since $C$ is closed, $X \setminus C$ is open in $X$. Consider the collection: $\{U_\alpha\}_{\alpha \in A} \cup \{X \setminus C\}.$

This is an open cover of $X$: For any $x \in X$, either $x \in C$ (in which case $x \in V_\alpha = U_\alpha \cap C$ for some $\alpha$, so $x \in U_\alpha$) or $x \notin C$ (so $x \in X \setminus C$).

Extract a finite subcover: Since $X$ is compact, there exist $\alpha_1, \ldots, \alpha_n \in A$ such that: $X = U_{\alpha_1} \cup \cdots \cup U_{\alpha_n} \cup (X \setminus C).$

Restrict to $C$: Intersecting both sides with $C$: $C = (U_{\alpha_1} \cap C) \cup \cdots \cup (U_{\alpha_n} \cap C) \cup \underbrace{((X \setminus C) \cap C)}_{= \emptyset} = V_{\alpha_1} \cup \cdots \cup V_{\alpha_n}.$

Therefore $\{V_{\alpha_1}, \ldots, V_{\alpha_n}\}$ is a finite subcover of the original cover of $C$, and $C$ is compact.

We use the finite intersection property characterization. Let $\{F_\beta\}_{\beta \in B}$ be a collection of closed subsets of $C$ (closed in the subspace topology) with the FIP. Since $C$ is closed in $X$, each $F_\beta$ is also closed in $X$.

The collection $\{F_\beta\}$ (viewed as subsets of $X$) has the FIP: any finite intersection $F_{\beta_1} \cap \cdots \cap F_{\beta_n} \neq \emptyset$ (this holds since FIP is the same regardless of the ambient space).

But wait: we need $\{F_\beta\}$ together with $C$ to have the FIP in $X$. Consider the collection $\{F_\beta\}_{\beta} \cup \{C\}$ in $X$. Each $F_\beta \subseteq C$, so $F_{\beta_1} \cap \cdots \cap F_{\beta_n} \cap C = F_{\beta_1} \cap \cdots \cap F_{\beta_n} \neq \emptyset$. This collection has the FIP.

Since $X$ is compact, by the FIP characterization: $\left(\bigcap_{\beta \in B} F_\beta\right) \cap C = \bigcap_{\beta \in B} F_\beta \neq \emptyset$ (the last equality since $F_\beta \subseteq C$).

So $\bigcap F_\beta \neq \emptyset$, and $C$ is compact by the FIP characterization.

We show $X \setminus K$ is open. Let $x \in X \setminus K$. For each $k \in K$, the Hausdorff property yields disjoint open sets $U_k \ni x$ and $V_k \ni k$. The collection $\{V_k\}_{k \in K}$ is an open cover of $K$. By compactness, there is a finite subcover $\{V_{k_1}, \ldots, V_{k_n}\}$.

Set $U = U_{k_1} \cap \cdots \cap U_{k_n}$. Then $U$ is open, $x \in U$, and $U \cap (V_{k_1} \cup \cdots \cup V_{k_n}) = \emptyset$. Since $K \subseteq V_{k_1} \cup \cdots \cup V_{k_n}$, we have $U \cap K = \emptyset$, i.e., $U \subseteq X \setminus K$.

Thus $X \setminus K$ is open, so $K$ is closed.

($\Rightarrow$): Compact in Hausdorff $\Rightarrow$ closed (Theorem 5.16).

($\Leftarrow$): Closed in compact $\Rightarrow$ compact (Theorem 5.15).