Assuming Choice, a topological group is profinite exactly when it is compact, Hausdorff, and totally disconnected

If $G$ is profinite, then by [F1] it is topologically isomorphic to an inverse limit of finite discrete groups. Therefore [L1] gives that $G$ is compact, Hausdorff, and totally disconnected. The same presentation gives the open normal basis in [L2]; each quotient by a coordinate kernel is isomorphic to the image of a projection and is therefore finite.

Conversely, assume that $G$ is compact, Hausdorff, and totally disconnected. By [L3], $G$ is totally separated. Let $U$ be an identity neighbourhood and choose an open identity neighbourhood $O\subseteq U$. For each $x\in G\setminus O$, total separatedness gives a clopen set $A_x$ containing $1$ but not $x$. The open complements $G\setminus A_x$ cover the compact set $G\setminus O$, so finitely many suffice. Their complementary intersection is a clopen identity neighbourhood $C\subseteq O\subseteq U$.

Put $K:=(G\setminus C)\cap C^2$. The set $C$ is compact by [L7], so $C\times C$ is compact by [L6], its product image $C^2$ is compact by [L5], and $K$ is compact by [L7]. Since $G$ is Hausdorff, [L5] makes $K$ closed. For each $h\in C$, continuity of multiplication at $(h,1)$ and openness of $C$ give open neighbourhoods $W_h\subseteq C$ of $h$ and $X_h\subseteq C$ of $1$ such that $W_h{X}_h\subseteq (G\setminus K)\cap C^2\subseteq C\mathrm{.}$ Compactness of $C$ gives finitely many $W_{h_j}$ covering it.

Let $X:={\bigcap }_j{X}_{h_j}$ and $Y:=X\cap X^{-1}$. Then $Y$ is a symmetric open identity neighbourhood and $CY\subseteq C$. Hence $H_0:={\bigcup }_{n\ge 1}{Y}^n$ is an open subgroup contained in $C$. Its open cosets cover compact $G$, so $H_0$ has finite index. The normalizer of $H_0$ contains $H_0$, so $H_0$ has only finitely many distinct conjugates. Their intersection $N$ is therefore an open normal finite-index subgroup with $N\subseteq H_0\subseteq C\subseteq U$. Since $U$ was arbitrary, these $N$ form an identity-neighbourhood basis.

Now let $G$ be any compact Hausdorff topological group having an identity-neighbourhood basis of open normal finite-index subgroups; this includes the situation produced in step 3.1. Let $\mathcal{N}$ be the family of all open normal finite-index subgroups. It still contains such a basis, and it is directed under reverse inclusion because it is closed under finite intersections. The quotient maps define a homomorphism Every coordinate map is continuous because its fibres are open cosets of $N$, so [L4] makes $\eta$ continuous.

The basis separates points: if $g\mathrm{\ne }1$, Hausdorffness gives an identity neighbourhood omitting $g$, and some $N\in \mathcal{N}$ lies inside it. Hence $\eta$ is injective. To prove density, let a nonempty basic cylinder in the inverse limit restrict the finitely many coordinates $N_1,\dots ,N_k$. If $k=0$, it contains $\eta (1)$. Otherwise put $M:=N_1\cap \cdots \cap N_k\in \mathcal{N}$, choose a compatible tuple in the cylinder, and choose a representative $g\in G$ of its $M$-coordinate. Compatibility makes the image of $g$ equal to the prescribed value in every restricted coordinate, so $\eta (g)$ lies in the cylinder. Thus $\eta [G]$ is dense.

By compactness of $G$ and [L5], the continuous image $\eta [G]$ is compact. The target is Hausdorff by [L1], so [L5] makes the image closed. It is both closed and dense by step 5.1, hence is the whole target. Thus $\eta$ is a continuous bijection from compact to Hausdorff and is a homeomorphism by [L5]. It is a group isomorphism by construction, so [F1] makes $G$ profinite.

Step 1.1 proves both forward implications. For a compact Hausdorff totally disconnected group, steps 1.2-3.1 construct the required basis and steps 4.1-6.1 make $G$ profinite. For a compact Hausdorff group already having the stated basis, steps 4.1-6.1 apply directly. This establishes both equivalences exactly as stated.