A nonempty compact connected one-dimensional manifold without boundary is a circle

Choose coordinate arcs with smaller closed coordinate intervals whose interiors cover $X$. Compactness gives finitely many such intervals $K_i$, each embedded in its coordinate chart. Their images are compact and closed by [F6]. Let $F$ be the finite set of all their endpoints; it is nonempty. In each $K_i$, cutting at its finitely many points of $F$ gives finitely many open intervals. Each such interval $C$ is open in $X$, connected by [F6], and closed in $X\setminus F$, because its compact closure is the corresponding embedded closed interval with its two endpoints in $F$. If two cut intervals meet, connectedness and this open-and-closed property force each to lie in the other, so they are equal. Every point of $X\setminus F$ lies in one of them, since it lies in some $K_i$ and is not an endpoint. Thus the distinct cut intervals form a finite partition of $X\setminus F$, each with an embedded closed-arc closure and two distinct endpoints in $F$.

At any $v\in F$, choose a sufficiently small coordinate interval containing no other point of $F$. Its two half-intervals lie in two incident cut-arc ends, and every incident arc approaching $v$ occupies one of these two sides. Hence exactly two ends meet at $v$. Start with one arc and follow its other endpoint by the unique other incident arc. Since there are finitely many vertices, a vertex repeats. The first repeated vertex is the starting vertex: a different earlier vertex already had both incident ends used on its first visit, so arrival from a previously unvisited vertex would require a third end. The resulting cyclic chain uses both ends at each of its vertices. Its union is closed, being a finite union of compact closed arcs, and open: interior points have interval neighbourhoods, and at its vertices both local sides belong to that union. It is nonempty, so connectedness of $X$ makes this cyclic chain all of $X$. This proves the cycle conclusion also when two different arcs have the same pair of endpoints.

Divide $S^1$ into the same finite number of consecutive closed angular arcs and map them, in cyclic order, onto the closed coordinate arcs of step 2.1, parametrizing each by its interval coordinate. Adjacent endpoints agree and only these endpoints are identified. Finite closed pasting gives a continuous bijection $S^1\to X$; [F6] makes it a homeomorphism. This proves the topological assertion without importing the classification of all connected topological one-manifolds.

If $X$ has a smooth structure, use the locally proved smooth classification [F4]. Its finitely many components are circles or closed intervals. Empty boundary excludes every interval; nonemptiness and connectedness leave exactly one circle. Thus the smooth assertion is a diffeomorphism with the standard circle. The finite topological construction used no extra choice; the countable-choice hypothesis is inherited from the smooth supplier. The empty manifold and the closed interval fail the respective explicit hypotheses, as stated.