A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism

For claim 1 assume $(X, \mathcal{T}_X)$ is compact, let $f : X \to Y$ be continuous, and let $\mathcal{U}$ be a family of open subsets of $Y$ with $f[X] \subseteq \bigcup \mathcal{U}$; put $\mathcal{G} := \{\, f^{-1}[U] : U \in \mathcal{U} \,\}$, a family of open subsets of $X$ by [L1], whose union is $X$ because every $x \in X$ has $f(x) \in U$ for some $U \in \mathcal{U}$.

If $X = \varnothing$ then $f[X] = \varnothing$ and the second alternative of [L3] holds; otherwise compactness of $X$ applied to $\mathcal{G}$ gives $n \in \mathbb{N}$ and $G_0, \dots, G_n \in \mathcal{G}$ with $X = G_0 \cup \dots \cup G_n$.

For each $j \le n$ the set $S_j := \{\, U \in \mathcal{U} : f^{-1}[U] = G_j \,\}$ is nonempty by the definition of $\mathcal{G}$, and $j \mapsto S_j$ is a function with domain the natural number $\sigma(n)$, so a choice function for its values supplies $U_0, \dots, U_n \in \mathcal{U}$ with $f^{-1}[U_j] = G_j$ for every $j \le n$.

Every point of $f[X]$ is $f(x)$ for some $x \in X$, and $x$ lies in some $G_j = f^{-1}[U_j]$, so $f(x) \in U_j$; hence $f[X] \subseteq U_0 \cup \dots \cup U_n$, and by [L3] the set $f[X]$ is a compact subset of $Y$.

For the second sentence of claim 1 let $K \subseteq X$ be a compact subset, so that the subspace $(K, (\mathcal{T}_X)_K)$ is a compact space by [L2] and $f|_K : K \to Y$ is continuous by [L5]; step 4.1, proved for an arbitrary compact space and an arbitrary continuous map out of it, applies to $f|_K$ and gives that $f|_K[K] = f[K]$ is a compact subset of $Y$.

For claim 2 assume $X$ is compact and nonempty and let $g : X \to \mathbb{R}$ be continuous; by step 4.1 the set $S := g[X]$ is a nonempty compact subset of $\mathbb{R}$. Suppose for the moment that $S$ has no maximum; then every $x \in S$ admits $s \in S$ with $x < s$, so the family $\mathcal{R} := \{\, (-\infty, s) : s \in S \,\}$ covers $S$, and its members are open by [L6], since $x < s$ gives $(x - r, x + r) \subseteq (-\infty, s)$ for $r := s - x > 0$.

By [L3] there are $n \in \mathbb{N}$ and $s_0, \dots, s_n \in S$ with $S \subseteq (-\infty, s_0) \cup \dots \cup (-\infty, s_n)$; by [L7] the set $\{s_0, \dots, s_n\}$ has a maximum, one of the $s_i$ and hence a member of $S$, so it lies in some $(-\infty, s_j)$, giving that maximum $< s_j$ while $s_j \le$ that maximum, which [L8] forbids. So $S$ has a maximum; the same argument with the rays $(s, \infty)$, open by [L6], and the minimum supplied by [L7] shows that $S$ has a minimum.

For claim 3 let $f : X \to Y$ be a continuous bijection with $X$ compact and $Y$ Hausdorff, and let $F \subseteq X$ be closed; then $F$ is a compact subset of $X$ by [L9], so $f[F]$ is a compact subset of $Y$ by step 5.1, and hence closed in $Y$ by [L10].

The maximum and the minimum of $S$ are members of $S = g[X]$, so there are $x_{\max}, x_{\min} \in X$ with $g(x_{\max})$ the maximum and $g(x_{\min})$ the minimum, and then $g(x_{\min}) \le g(x) \le g(x_{\max})$ for every $x \in X$; this is claim 2.

Step 6.2 says that $f$ carries closed sets to closed sets, so $f$ is a closed map, and by [L11] a continuous bijection that is closed is a homeomorphism, which is claim 3; claims 1 and 2 were proved at steps 5.1 and 7.1.