A metric space $(X,d)$ is said to be a geodesic metric space if and only if each pair of points $x,y \in X$ is connected by a geodesic, where the geodesic need not be unique.
I assume that having this property does not necessarily need the space also needs to be a length space, but I have seen mention that a geodesic space is a length space to begin with by definition.
Let $(X, d)$ be a geodesic space, and let $x, y\in X$. Let $\sigma : [0,1]\to X$ be any rectifiable curve joining $x, y$. Then
$$L(\sigma) \ge d(x, y)$$
by definition of $L(\sigma)$. On the other hand, let $\gamma : [0,d]\to X$ be a geodesic joining $x, y$, where $d = d(x,y)$. By definition, $$d(\gamma(s), \gamma(t)) = v|s-t|$$ for all $s, t\in [0,d]$. Putting $s=0, t = d$, we have $v = 1$. Also,
$$L(\gamma) = \sup \sum_{i=1}^n d(\gamma(t_{i-1}, t_i) = \sup \sum_{i=1}^n (t_i - t_{i-1}) = d.$$
(the supremum is taken over all partitions of $[0,d]$) Thus $\gamma$ is rectifiable and $L(\gamma) = d(x, y)$. Hence
$$ \inf_{\sigma\ \ \text{rectifiable}} L(\sigma) = L(\gamma) = d(x, y)$$
and thus $(X, d)$ is a length space.