It is well-known, that a complete metric space, where any two points have a midpoints ($\forall x,y~ \exists z:~d(x,z)=d(y,z)=\frac{d(x,y)}{2}$) is strictly intrinsic, in the sense that any $x,y$ can be joined by a path of length $d(x,y)$.

Also, completeness and existence of $\varepsilon$-midpoints ($\forall x,y~ \exists z:~|d(x,z)-\frac{d(x,y)}{2}|+|d(y,z)-\frac{d(x,y)}{2}|\le \varepsilon$) for any $\varepsilon$ imply the space being intrinsic, i.e. any $x,y$ can be joined by a path of length $d(x,y)+\varepsilon$ for any $\varepsilon$.

Can we replace completeness by something else?

It seems that $\varepsilon$-midpoints are completely useless without completeness (a disc without one radius is an example).

On the other hand, it seems that the existence of midpoints is a rather strong condition. So far I have the following examples:

$\mathbb{R}\times(0,+\infty)\bigcup \mathbb{Q}$ with the induced metric is a space with midpoints, which is not strictly intrinsic, but still it is intrinsic.

$\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\{x=y\}$ with the induced $L_{\infty}$ metric is a connected space with midpoints but not intrinsic. 

However these space are bad: neither of them is locally compact, and the latter is not even locally path connected  (local compactness implies local completeness and so local strict intrinsicness, and so local path connectedness).

Now let us assume that in connected locally compact (or merely locally path connected) metric space $X$ for any two points there is a midpoint. Is it true that this space is (strictly) intrinsic?

There were numerous edits, because I have added two counterexamples, one of which later proved to be completely incorrect, and another one had a mistake.