It seems that the answer is yes.

We may replace $\alpha(d)$ by $\beta(d)=\sup_{t\geq d}\alpha(t)$, thus obtaining that $\alpha(d)$ is stricty monotone.

As usual, start with any $x_0\in X$ and define $x_{n+1}=f(x_n)$, $d_{n+1}=d(x_n,x_{n+1})\leq \alpha(d_n)d_n$. The monotone sequence $d_n$ converges to some $D$; if $D>0$ then we have $D\leq \alpha(D)D$ which is absurd. Hence $D=0$.

Now denote $r_n=\sup_{k>n}d(x_n,x_k)$. We want to show that $r_n\to 0$ as $n\to 0$ (in particular, this will yield that all the $r_n$ are finite). Assume, to the contrary, that there exists some $\mu>0$ such that $r_n>\mu$ infinitely often.

Set $\nu=\alpha(\mu)$. There exists an $n$ with $r_n>\mu$ such that $d_n<\frac{(1-\nu)\mu}{2\nu}$. Choose $k>n$ such that $d(x_n,x_k)>\mu$ (then $d(x_{n-1},x_{k-1})>\mu$ as well). Then we have $$ d(x_n,x_k)\leq \alpha(d(x_{n-1},x_{k-1}))d(x_{n-1},x_{k-1}) \leq \nu(d_n+d(x_n,x_k)+d_k) \leq \nu(d(x_n,x_k)+2d_n), $$ or $(1-\nu)\mu<(1-\nu)d(x_n,x_k)\leq 2\nu d_n$. This contradicts the choice of $n$.

Thus $r_n$ indeed tend to $0$; therefore, our sequence $(x_n)$ is convergent to the common point of the balls $B_{r_n}(x_n)$. Surely, this limit is the sought fixed point.

It seems that the answer is yes.

We may replace $\alpha(d)$ by $\beta(d)=\sup_{t\geq d}\alpha(t)$ (surely, $\beta(q)$ is the limit of some sequence $\alpha(t_n)$ with $t_n\geq d$, so $\beta(d)<1$). Thus we obtain monotone $\alpha(d)$.

As usual, start with any $x_0\in X$ and define $x_{n+1}=f(x_n)$, $d_{n+1}=d(x_n,x_{n+1})\leq \alpha(d_n)d_n$. The monotone sequence $d_n$ converges to some $D$; if $D>0$ then we have $D\leq \alpha(D)D$ which is absurd. Hence $D=0$.

Now denote $r_n=\sup_{k>n}d(x_n,x_k)$. We want to show that $r_n\to 0$ as $n\to 0$ (in particular, this will yield that all the $r_n$ are finite). Assume, to the contrary, that there exists some $\mu>0$ such that $r_n>\mu$ infinitely often.

Set $\nu=\alpha(\mu)$. There exists an $n$ with $r_n>\mu$ such that $d_n<\frac{(1-\nu)\mu}{2\nu}$. Choose $k>n$ such that $d(x_n,x_k)>\mu$ (then $d(x_{n-1},x_{k-1})>\mu$ as well). Then we have $$ d(x_n,x_k)\leq \alpha(d(x_{n-1},x_{k-1}))d(x_{n-1},x_{k-1}) \leq \nu(d_n+d(x_n,x_k)+d_k) \leq \nu(d(x_n,x_k)+2d_n), $$ or $(1-\nu)\mu<(1-\nu)d(x_n,x_k)\leq 2\nu d_n$. This contradicts the choice of $n$.

Thus $r_n$ indeed tend to $0$; therefore, our sequence $(x_n)$ is convergent to the common point of the balls $B_{r_n}(x_n)$. Surely, this limit is the sought fixed point.

It seems that the answer is yes.

We may replace $\alpha(d)$ by $\beta(d)=\sup_{y\geq d}\alpha(t)$, thus obtaining that $\alpha(d)$ is stricty monotone.

As usual, start with any $x_0\in X$ and define $x_{n+1}=f(x_n)$, $d_{n+1}=d(x_n,x_{n+1})\leq \alpha(d_n)d_n$. The monotone sequence $d_n$ converges to some $D$; if $D>0$ then we have $D\leq \alpha(D)D$ which is absurd. Hence $D=0$.

Now denote $r_n=\sup_{k>n}d(x_n,x_k)$. We want to show that $r_n\to 0$ as $n\to 0$ (in particular, this will yield that all the $r_n$ are finite). Assume, to the contrary, that there exists some $\mu>0$ such that $r_n>\mu$ infinitely often.

Set $\nu=\alpha(\mu)$. There exists an $n$ with $r_n>\mu$ such that $d_n<\frac{(1-\nu)\mu}{2\nu}$. Choose $k>n$ such that $d(x_n,x_k)>\mu$ (then $d(x_{n-1},x_{k-1})>\mu$ as well). Then we have $$ d(x_n,x_k)\leq \alpha(d(x_{n-1},x_{k-1}))d(x_{n-1},x_{k-1}) \leq \nu(d_n+d(x_n,x_k)+d_k) \leq \nu(d(x_n,x_k)+2d_n), $$ or $(1-\nu)\mu<(1-\nu)d(x_n,x_k)\leq 2\nu d_n$. This contradicts the choice of $n$.

Thus $r_n$ indeed tend to $0$; therefore, our sequence $(x_n)$ is convergent to the common point of the balls $B_{r_n}(x_n)$. Surely, this limit is the sought fixed point.

It seems that the answer is yes.

We may replace $\alpha(d)$ by $\beta(d)=\sup_{t\geq d}\alpha(t)$, thus obtaining that $\alpha(d)$ is stricty monotone.

As usual, start with any $x_0\in X$ and define $x_{n+1}=f(x_n)$, $d_{n+1}=d(x_n,x_{n+1})\leq \alpha(d_n)d_n$. The monotone sequence $d_n$ converges to some $D$; if $D>0$ then we have $D\leq \alpha(D)D$ which is absurd. Hence $D=0$.

Now denote $r_n=\sup_{k>n}d(x_n,x_k)$. We want to show that $r_n\to 0$ as $n\to 0$ (in particular, this will yield that all the $r_n$ are finite). Assume, to the contrary, that there exists some $\mu>0$ such that $r_n>\mu$ infinitely often.

Set $\nu=\alpha(\mu)$. There exists an $n$ with $r_n>\mu$ such that $d_n<\frac{(1-\nu)\mu}{2\nu}$. Choose $k>n$ such that $d(x_n,x_k)>\mu$ (then $d(x_{n-1},x_{k-1})>\mu$ as well). Then we have $$ d(x_n,x_k)\leq \alpha(d(x_{n-1},x_{k-1}))d(x_{n-1},x_{k-1}) \leq \nu(d_n+d(x_n,x_k)+d_k) \leq \nu(d(x_n,x_k)+2d_n), $$ or $(1-\nu)\mu<(1-\nu)d(x_n,x_k)\leq 2\nu d_n$. This contradicts the choice of $n$.

Thus $r_n$ indeed tend to $0$; therefore, our sequence $(x_n)$ is convergent to the common point of the balls $B_{r_n}(x_n)$. Surely, this limit is the sought fixed point.