Such a manifold has bounded diameter, and is therefore compact. To see this, let $0 < T < \infty$ be the quantity you define. I claim that $\mathrm{diam} M \leq T$.
Additionally for each point $p \in M$ and unit tangent vector $v \in T_p M$ let $\tau(v)$ be the first conjugate time along the geodesic $\gamma: t \mapsto \exp_p(tv)$. Explicitly, $\tau(v) > 0$ is the first time $t > 0$ so that $\gamma(t)$ is a conjugate point of $p$ along the geodesic.
Then $\tau(v) \leq T$ for all unit vectors, which can be obtained as follows. Let $\rho_j \downarrow 0$ and $K_j = \{ w \in T_p M \mid \lvert w \rvert = 1, \lvert w - v \rvert \leq \rho_j \}$. By definition of $T$ there are vectors $v_j \neq w_j \to v$ so that the corresponding geodesics intersect before this time. Therefore $\gamma$ has a conjugate point before or at $\gamma(T)$.
It is well-known that a geodesic is not length-minimising past its first conjugate point. Therefore every length-minimising geodesic has length at most $T$, and the claim is proved.
Edit. To see why $\gamma$ must reach a conjugate point before time $T$, one can argue as follows. Let $q_j \neq p$ be the first intersection point of the geodesics directed by $v_j,w_j$. Because $v_j,w_j \to v$ we may extract a subsequence so that $q_j \to q$ as $j \to \infty$, where $q = \gamma(t)$ for some $t > 0$. This point $q$ is conjugate to $p$. If $D \exp_p(t v)$ were non-singular then there would exist a neighbourhood $U \subset T_p M$ of $tv$ so that $\exp_p \mid U$ is a diffeomorphism onto its image. But eventually $t v_j \neq t w_j$ both lie in $U$, and $q_j$ in its image. In particular $\exp_p$ is not injective in $U$.
