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. A better example: $[-1,1]\times[0,1]\backslash\{0\}$ with the induced $L_{\infty}$ metric is even a locally compact space with midpoints, which is not strictly intrinsic. These spaces however are intrinsic.

On the other hand the space $\{(x,y)\in \mathbb{R}^{2}, x-y\in\mathbb{Q}\}\bigcup\{x=y\}$ with the induced $L_{\infty}$ metric is a connected space with midpoints but not intrinsic. However this is a bad space: it is neither locally path connected, nor locally compact.

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 intrinsic?

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

EDIT: I have added an example of a locally compact connected locally path connected space with midpoints but not strictly intrinsic and an example of a connected space with midpoints but not intrinsic. These counterexamples refined the original two questions into their "maximum".