= Solution
A <topological graph> is a <CW complex> with only zero-dimensional and one-dimensional cells: its vertices form the zero-skeleton, and each edge is a copy of an open interval attached by its two endpoints to vertices, possibly the same vertex. It has the <weak topology of a CW complex>. A <subgraph> is a <CW subcomplex>, obtained by choosing vertices and edges and retaining the endpoints of every chosen edge. Loops and multiple edges are allowed.
A <spanning tree> is a connected <spanning subgraph> containing no <cycle in a graph>. To construct one in any connected <topological graph> $\Gamma$, consider all acyclic spanning subgraphs, ordered by inclusion; the subgraph with every vertex and no edges is one. The union of a chain is still acyclic, since any cycle uses only finitely many edges and would already occur in a member of the chain. The <Zorn lemma> therefore gives a maximal such subgraph $\Delta$.
If $\Delta$ were disconnected, a finite edge path in $\Gamma$ between vertices of two of its components would contain an edge joining different components. Adding that edge creates no cycle, contradicting maximality. Thus $\Delta$ is a <spanning tree>. Fix a root vertex. Each point of $\Delta$ has a unique finite path to the root; give every edge length one and move the point a fraction $s$ of the way along that path at time $s\in[0,1]$. This defines a <contraction of a topological space> of $\Delta$. On each closed edge times $[0,1]$, the construction takes place in a finite tree and is continuous. Since the interval is a locally finite <CW complex>, the product with $\Delta$ has the corresponding <weak topology of a CW complex>, so these restrictions establish continuity globally. Consequently
$$
\boxed{\Delta\simeq\{\text{point}\}.}
$$
This argument includes infinite graphs and does not assume local finiteness of $\Gamma$.
If $p:\widetilde\Gamma\to\Gamma$ is a <covering map>, take $p^{-1}(\Gamma^0)$ as the vertices upstairs. Each component of the inverse image of an open edge is an open interval mapping homeomorphically to that edge, because an interval is <simply connected>. Its parametrization lifts along the closed edge to attach its endpoints to lifted vertices, by <path lifting theorem>. Small star neighborhoods at vertices are lifted homeomorphically by the <covering map>, and give exactly the neighborhoods of the resulting <topological graph>. Thus the cell structure has the topology of the original covering space. The total space may be disconnected, but every component is a <topological graph>.
The <Nielsen–Schreier formula> is
$$
\boxed{[F_n:H]=d<\infty\quad\Longrightarrow\quad H\cong F_{1+d(n-1)}.}
$$
For a proof, realize $F_n$ as the <fundamental group> of the <wedge sum> of $n$ circles. Construct its connected <covering graph> with vertices the right <cosets> $Hg$, and an oriented edge labeled $x_i$ from $Hg$ to $Hgx_i$ for each free generator $x_i$. There is exactly one incoming and one outgoing edge of each label at every vertex, which establishes the local <covering map> property. Reading a word from $H$ ends at its coset, so the closed paths at $H$ give exactly the words in $H$. The induced map of <fundamental groups> is injective by the <homotopy lifting property> of a <covering map>, and its image is $H$.
For index $d$, this <covering graph> has $d$ vertices and $dn$ edges. A <spanning tree> in it has $d-1$ edges. The permitted rank formula for a finite connected graph now gives
$$
\operatorname{rank}H=dn-(d-1)=1+d(n-1).
$$
In particular this proves both freeness and the index formula.
If $n\ge2$ and $d>1$, then $1+d(n-1)>n$, so a proper finite-index <subgroup> cannot be isomorphic to $F_n$. For $n=1$, $F_1\cong\mathbb Z$, and $d\mathbb Z$ is a proper index-$d$ <subgroup> isomorphic to $\mathbb Z$ for every $d\ge2$. Hence
$$
\boxed{\text{Such a proper finite-index copy exists exactly when }n=1.}
$$
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