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 $X=\mathrm{\varnothing }$ then $f[X]=\mathrm{\varnothing }$ 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 $X=G_0\cup \cdots \cup G_n$.

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 $f[X]\subseteq U_0\cup \cdots \cup U_n$, 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 $f{\mathrm{\mid }}_K:K\to Y$ is continuous by [L5]; step 4.1, proved for an arbitrary compact space and an arbitrary continuous map out of it, applies to $f{\mathrm{\mid }}_K$ and gives that $f{\mathrm{\mid }}_K[K]=f[K]$ 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 $\mathcal{R}:=\{(-\mathrm{\infty },s):s\in S\}$ covers $S$, and its members are open by [L6], since $x<s$ gives $(x-r,x+r)\subseteq (-\mathrm{\infty },s)$ for $r:=s-x>0$.

By [L3] there are $n\in \mathbb{N}$ and $s_0,\dots ,s_n\in S$ with $S\subseteq (-\mathrm{\infty },s_0)\cup \cdots \cup (-\mathrm{\infty },s_n)$; 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 $(-\mathrm{\infty },s_j)$, 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 $(s,\mathrm{\infty })$, 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.