Here are my thoughts so far:
In what follows, curves $\gamma$ are always weak contractions. Lemma 0 and 1 are general remarks about length spaces.
Lemma 0. If $\gamma\colon[0,L]\to X$ is a weak contraction with $x=\gamma(0)\ne \gamma(L)=y$, then we may assume wlog. that $\gamma$ is injective.
Proof:
Let $S$ be the set of all finite unions of intervals of the form
$$ A=\bigcup_{k=1}^n[a_k,b_k]$$
with $a_1=0$, $b_n=L$, $a_k<b_k<a_{k+1}$, $\gamma(b_k)=\gamma(a_{k+1})$.
For $A\in S$ we have $\mu(A)=\sum (b_k-a_k)\ge d(x,y)$ and can define the weak contraction
$$\gamma_A\colon[0,L-\mu(A)]\to X , t\mapsto \gamma(\mu(A\cap[0,t])).$$
Let $(A_n)_n$ be a sequence in $S$ such that $\mu(A_n)$ converges to $m:=\sup\{\,\mu(A)\mid A\in S\,\}$. One checks that $(\gamma_{A_n}|_{[0,L-m]})_n$ converges pointwise to a weak contraction $[0,L-m]\to X$ that is injective.$_\square$
Lemma 1. Every open ball is (path-)connected.
Proof: Indeed, If $y\in B_r(x)$, then there is a curve $\gamma$ from $x$ to $y$ that stays within $B_r(x)$, e.g. we can take $L=\frac{r+d(x,y)}2$.$_\square$
Lemma 2. If $\gamma$ is a curve from $x$ to $y\ne x$ with $L<d(x,y)+\epsilon$, then for every $t\in(3\epsilon,L-3\epsilon)$ there is a neighbourhood of $\gamma(t)$ that is homeomorphic to an open interval.
Proof: Let
$$\tag1 U=\bigcup_{0<t<L}B_{\min\{\epsilon,t,L-t\}}(\gamma(t)).$$
As union of open sets, $U$ is open. Moreover, the single balls are connected by lemma 1 and are overlapping. This makes $U$ connected.
For some $t_0\in(3\epsilon,L-3\epsilon)$, consider $$\tag2U'=U\setminus\{z\}\quad\text{with }z=\gamma(t_0).$$
By assumption, $U'$ can be written as disjoint union of open balls.
If we assume that $U'$ is still connected, this union is in fact just a single ball $B_r(\xi)$.
Because $U'$ overlaps every ball around $x$ (or $y$ or $z$), we see that $d(\xi,x)=d(\xi,y)=d(\xi,z)=r$, hence $$\tag32r=d(\xi,x)+d(\xi,y)\ge d(x,y)>L-\epsilon.$$
Since there is a $\tau\in[0,L]$ with $d(z,\gamma(\tau))<\epsilon$, we have $$\tag4r=d(\xi,x)<\tau+\epsilon$$ and $$\tag5r=d(\xi,y)<(L-\tau)+\epsilon,$$ hence $r<\frac L2+\epsilon$. Also, $$\tag6r=d(\xi,z)<\epsilon+|\tau-t_0|.$$
If $t_0\le \tau$, this implies $r<\epsilon+\tau-t_0$ and together with
(5) and (3)
$$ (L-t_0)+2\epsilon>2r>L-\epsilon.$$
If on the other hand $t_0>\tau$, then (6), (4) and (3) impliy
$$ t_0+2\epsilon>2r>L-\epsilon.$$
In both cases we obtain a contradiciton to $t_0\in(3\epsilon,L-3\epsilon)$.
Therefore, $U'$ is not connected.
Now for a point $\eta\in U'$, the following cases are possible:
$\eta\in B_\epsilon(z)$. Then also $\eta\in B_\epsilon(z')$ for some $z'=\gamma(t')$ with $t'\approx t_0$. Therefore there is a path within $U'$ from $\eta$ to $\gamma(t')$ and from there along $\gamma$ to either $x$ or $y$.
$\eta\notin B_\epsilon(z)$. Then of course $\eta\in B_{\min\{\epsilon,t',L-t'\}}(\gamma(t'))$ for some $t'\ne t_0$ and again $\eta$ is path-connected within $U'$ to $x$ or $y$.
We conclude that $U'$ has exactly two connected components, one containing $x$ and one containing $y$.
Let $\delta =\min\{\epsilon,t_0-3\epsilon,L-3\epsilon-t_0\}>0$.
Assume that $B_\delta(z)$ contains a point $\eta$ that is not in the image of $\gamma$.
For $t_1$ sufficiently close to $t_0$ (and with $t_0<t_1<t_0+\delta$, say), the point $\eta$ is also in $B_\epsilon(\gamma(t_1))$.
By the same argument as above, in the set $U''=U\setminus\{\gamma(\frac{t_0+t_1}2)\}$ the points $x$ and $y$ are in different connected components. But in fact they are connected via paths $x\to \gamma(t_0)\to\eta\to\gamma(t_1)\to y$.
Therefore $B_\delta(z)$ contains only points in $\gamma([0,L])$ and is homeomorphic image of a connected subset of $[0,L]$.$_\square$
Lemma 3. If $x\ne y$ then there is a curve $\gamma\colon[0,L]$ from $x$ to $y$ such that $\gamma|_{(0,L)}$ is a homeomorphism with an open subset of $X$.
Proof:
Using lemma 2 with $\epsilon<\frac16d(x,y)$ we find a point $z=\gamma(\frac L2)$ that has a neighbourhod looking like an interval. Any (injective) curve from $z$ to $x$ taking the "wrong" direction has a length of at least $\frac L2+\epsilon$, hence this does not happen for all sufficiently short curves from $z$ to $x$. We conclude that a sequence of curves from $z$ to $x$ where $L\to d(z,x)$ converges pointwise to a curve $[0,d(z,x)]\to X$ such that the restriction to $[0,d(z,x))$ is a homeomorphism. If we do the same with $y$ instead of $x$ and combine the results, the claim of the lemma follows.$_\square$
