Induct on $n = r + g + b$. The base case, $n = 1$, is clear.

Let now $n > 1$. Let $V$ denote the vertex set of $\Gamma$, and let $T_r, T_g$, and $T_b$ be the trees with exactly $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. Consider two cases.

Case 1: There exists a partition $V = A \cup B$ of the vertex set into two non-empty parts such that the edges joining the parts all bear the same colour, say, blue.

Since $\Gamma$ is connected, it has a (necessarily blue) edge connecting $A$ and $B$. Let $e$ be one such.

Assume that $T$, one of the three trees, does not contain $e$. Then the graph $T \cup \{e\}$ has a cycle $C$ through $e$. The cycle $C$ should contain another edge $e'$ connecting $A$ and $B$; the edge $e'$ is also blue. Replace $e'$ by $e$ in $T$ to get another tree $T'$ with the same number of edges of each colour as in $T$, but containing $e$.

Performing such an operation to all three trees, we arrive at the situation where the three trees $T'_r, T'_g$, and $T'_b$ all contain $e$. Now shrink $e$ by identifying its endpoints to obtain a graph $\Gamma^*$, and set $r^* = r, g^* = g$, and $b^* = b - 1$. The new graph satisfies the conditions in the statement for those new values — indeed, under the shrinking, each of the trees $T'_r, T'_g$, and $T'_b$ loses a blue edge. So $\Gamma^*$ has a spanning tree with exactly $r$ red, exactly $g$ green, and exactly $b - 1$ blue edges. Finally, pass back to $\Gamma$ by restoring $e$, to obtain the desired spanning tree in $\Gamma$.

Case 2: There is no such a partition.

Consider all possible collections $(R, G, B)$, where $R, G$ and $B$ are acyclic sets consisting of $r$ red edges, $g$ green edges, and $b$ blue edges, respectively. By the problem assumptions, there is at least one such collection. Amongst all such collections, consider one such that the graph on $V$ with edge set $R \cup G \cup B$ has the smallest number $k$ of components. If $k = 1$, then the collection provides the edges of a desired tree (the number of edges is one less than the number of vertices).

Assume now that $k \ge 2$; then in the resulting graph some component $K$ contains a cycle $C$. Since $R, G$, and $B$ are acyclic, $C$ contains edges of at least two colours, say, red and green. By assumption, the edges joining $V(K)$ to $V \setminus V(K)$ bear at least two colours; so one of these edges is either red or green. Without loss of generality, consider a red such edge $e$.

Let $e'$ be a red edge in $C$ and set $R' = R \setminus \{e'\} \cup \{e\}$. Then $(R', G, B)$ is a valid collection providing a smaller number of components. This contradicts minimality of the choice above and concludes the proof.

For a spanning tree $T$ in $\Gamma$, denote by $r(T)$, $g(T)$, and $b(T)$ the number of red, green, and blue edges in $T$, respectively.

Assume that $C$ is some collection of spanning trees in $\Gamma$. Write
$$r(C) = \min_{T \in C} r(T), \quad g(C) = \min_{T \in C} g(T), \quad b(C) = \min_{T \in C} b(T), \\ R(C) = \max_{T \in C} r(T), \quad G(C) = \max_{T \in C} g(T), \quad B(C) = \max_{T \in C} b(T).$$
Say that a collection $C$ is good if $r \in [r(C), R(C)]$, $g \in [g(C), G(C)]$, and $b \in [b(C), B(C)]$. By the problem conditions, the collection of all spanning trees in $\Gamma$ is good.

For a good collection $C$, say that an edge $e$ of $\Gamma$ is suspicious if $e$ belongs to some tree in $C$ but not to all trees in $C$. Choose now a good collection $C$ minimizing the number of suspicious edges. If $C$ contains a desired tree, we are done. Otherwise, without loss of generality, $r(C) < r$ and $G(C) > g$.

We now distinguish two cases.

Case 1: $B(C) = b$.

Let $T^0$ be a tree in $C$ with $g(T^0) = g(C) \le g$. Since $G(C) > g$, there exists a green edge $e$ contained in some tree in $C$ but not in $T^0$; clearly, $e$ is suspicious. Fix one such green edge $e$.

Now, for every $T$ in $C$, define a spanning tree $T_1$ of $\Gamma$ as follows. If $T$ does not contain $e$, then $T_1 = T$; in particular, $(T^0)_1 = T^0$. Otherwise, the graph $T \setminus \{e\}$ falls into two components. The tree $T^0$ contains some edge $e'$ joining those components; this edge is necessarily suspicious. Choose one such edge and define $T_1 = T \setminus \{e\} \cup \{e'\}$.

Let $C_1 = \{T_1 : T \in C\}$. All edges suspicious for $C_1$ are also suspicious for $C$, but no tree in $C_1$ contains $e$. So the number of suspicious edges for $C_1$ is strictly smaller than that for $C$.

We now show that $C_1$ is good, reaching thereby a contradiction with the choice of $C$. For every $T$ in $C$, the tree $T_1$ either coincides with $T$ or is obtained from it by removing a green edge and adding an edge of some colour. This already shows that $g(C_1) \le g(C) \le g$, $G(C_1) \ge G(C) - 1 \ge g$, $R(C_1) \ge R(C) \ge r$, $r(C_1) \le r(C) + 1 \le r$, and $B(C_1) \ge B(C) \ge b$. Finally, we get $b(T^0) \le B(C) = b$; since $C_1$ contains $T^0$, it follows that $b(C_1) \le b(T^0) \le b$, which concludes the proof.

Case 2: $B(C) > b$.

Consider a tree $T^0$ in $C$ satisfying $r(T^0) = R(C) \ge r$. Since $r(C) < r$, the tree $T^0$ contains a suspicious red edge. Fix one such edge $e$.

Now, for every $T$ in $C$, define a spanning tree $T_2$ of $\Gamma$ as follows. If $T$ contains $e$, then $T_2 = T$; in particular, $(T^0)_2 = T^0$. Otherwise, the graph $T \cup \{e\}$ contains a cycle $C$ through $e$. This cycle contains an edge $e'$ absent from $T^0$ (otherwise $T^0$ would contain the cycle $C$), so $e'$ is suspicious. Choose one such edge and define $T_2 = T \setminus \{e'\} \cup \{e\}$.

Let $C_2 = \{T_2: T \in C\}$. All edges suspicious for $C_2$ are also suspicious for $C$, but all trees in $C_2$ contain $e$. So the number of suspicious edges for $C_2$ is strictly smaller than that for $C$.

We now show that $C_2$ is good, reaching again a contradiction. For every $T$ in $C$, the tree $T_2$ either coincides with $T$ or is obtained from it by removing some edge and adding a red edge. This shows that $r(C_2) \le r(C) + 1 \le r$, $R(C_2) \ge R(C) \ge r$, $G(C_2) \ge G(C) - 1 \ge g$, $g(C_2) \le g(C) \le g$, $b(C_2) \le b(C) \le b$ and $B(C_2) \ge B(C) - 1 \ge b$. This concludes the proof.