All geodesics are inextendable? I think the title is true, because geodesics has a tangent vector with a constant length parametrized by an affine parameter. 
Probably, it is easier to think about timelike or spacelike geodesics. In this case, its affine parameter measures the length of the curve. It is difficult to image such kind of curve has a future endpoint (or past endpoint). 
Is it correct? 
 A: This is not true.  There are trivial counter-examples.
For example, take $\mathbb{R}$ with the trivial metric.  Then, $\gamma :(0,1)\rightarrow \mathbb{R}$ defined by $\gamma (t):=t$ is a geodesic, but it is also clearly extendable.  Indeed, it is extended by $\tilde{\gamma}:\mathbb{R}\rightarrow \mathbb{R}$ defined by $\tilde{\gamma}(t):=t$.
A: It is not true. Sometimes the spacetime itself and therefore the geodesics can be extended. Consider for example the manifold described by (t,x,y,z) with $x,y,z> 0$ and Minkowski metric. This is nothing more than Minkowski spacetime truncated to a smaller region, but it is a perfectly valid manifold for all purposes.

In the diagram we see schematically what this would look like in the x-y plane and the time direction not plotted. A possible geodesic is also depicted. It is easy to convince oneself that it is possible for this geodesic to be described by a parameter in the interval $\lambda\in (0,\infty)$, with $\lambda(0)=0$ and $\lambda(\infty)=\infty$.
We can then extend the initial spacetime to be the full Minkowski spacetime described by (t,x,y,z) with $x,y,z\in\mathbb R$ and Minkowski metric. In the new spacetime, we can also extend the geodesics to something bigger. In particular, the depicted geodesic would trivially extend to a geodesic described by the same parametric equation but in this case we would have $\lambda\in (-\infty,\infty)$.
This example might be fairly trivial, because you can immediately see that extending the range of coordinates, naturally extends the spacetime. However, it is possible to come up with spacetimes described by coordinates all of which are in the $(-\infty,\infty)$ and that can still be extended. This can be seen by first converting to different coordinates in some smaller range and then extending the spacetime using these new coordinates. In general, one has to be very careful to disentangle properties of the manifold itself form properties of any particular coordinate system used.
A: The property you are referring to is called geodesic completeness.  It is an important concept in the study of singularities in general relativity.  There are somewhat trivial examples of geodesic incompleteness where you are just "missing" part of the spacetime, i.e. you could have Minkowski space with a point removed.  In these cases you usually consider extending the spacetime ("isometrically embedding" it in a bigger one, without the "hole").  
The real interesting thing is when your spacetime has singularities.  For example, the maximally extended Schwarzschild black hole spacetime has a singularity at $r=0$, and geodesics run into it within finite affine parameter.  These geodesics cannot be extended.  
The proof of the singularity theorems of Penrose and Hawking basically amounts to proving that a spacetime is geodesically incomplete.  Hence, geodesic incompleteness can be taken as the definition of a spacetime with a singularity.  
A: A geodesic $\gamma$ is extendible only if $\gamma$ is incomplete, in the realm of semi-Riemaniann manifolds.
Suppose $\gamma$ is complete and extendible, say $\gamma:(-\infty,\infty)\to M$. Suppose $p$ is a future endpoint without lose of generality. Then, there is $t_0\in\mathbb{R}$, for all $t\ge t_0$, such that $\gamma(t)\in U_p$, where $U_p$ denotes an arbitrary neighborhood of $p$ in $M$. Let $U_p=\{x\in M:d(x,p)<\varepsilon\}$, where $d$ is the distance function defined in [O'Neill, Semi-Riemaniann Geometry, Definition 15.5]. Then, as we assumed, $\gamma(t_0)\in U_p$ and $\gamma(t_0+2\varepsilon)\in U_p$, and hence,
$$ d(\gamma(t_0+2\varepsilon),\gamma(t_0))\le d(\gamma(t_0+2\varepsilon),p)+d(\gamma(t_0),p)<2\varepsilon,$$
which contradicts with that $\gamma$ is a geodesic. 
As a result, $\gamma$ could not be both complete and extendible. 
