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

Now let us assume that in connected (even locally path-connected) metric space $X$ for any two points there is a midpoint. Is it true that this space is intrinsic? If we further assume say local compactness, is it true that the space is strictly intrinsic?

Local compactness implies local completeness and so local strict intrinsicness, but I don't know how to proceed.

My idea of potential counterexample is a space with a lack of rectifiable curves, but I cannot make one satisfy the midpoint condition.

Thank you.