We will show that if $(1)$ and $(3)$ fail, then $(2)$ must hold. In fact, it is possible to show a slightly stronger result: that if $\chi(G) \geq r + 1$ and there exists a tree $T$ on $r + 1$ vertices such that $G$ does not contain an induced copy of $T$, then $G$ must contain an induced cycle on at most $r + 1$ vertices.
We will need two lemmas. Recall that $\delta(G)$ denotes the minimum degree of a graph $G$.
Lemma 1. Any graph $G$ satisfies $\chi(G) \leq \max\{\delta(H) \mid H \subseteq G\} + 1.$
Lemma 2. If $\delta(G) \geq r$ and $T$ is a tree on $r + 1$ vertices, then $G$ contains a copy of $T$.
Now we are ready to begin.
Proof: Fix $r \in \mathbb{N}$ and suppose that $G$ is not $r$-colorable. By Lemma 1, $\chi(G) \geq r + 1$ implies that $\max\{\delta(H) \mid H \subseteq G\} \geq r$, that is, that $G$ has a subgraph $H$ of minimum degree at least $r$. By Lemma 2, $H$, and hence $G$, contains a copy of every tree on $r + 1$ vertices.
Suppose that there is some tree $T$ on $r + 1$ vertices such that $G$ does not contain an induced copy of $T$. Then $G$ must contain a cycle on at most $r + 1$ vertices. Furthermore, the (not necessarily unique) smallest such cycle must be induced. This completes the proof.