My Question
Is there a problem with this proof(red underlined part)? I do apologize for the use of images.
I believe that $\alpha$ is allowed to be discontinuous at $v_i$ and $u_i$ as we are only concerned about continuity at $d_i$ points.
The Theorem
Counter example
The equation $\alpha(t) - \alpha(d_i) < \frac{\epsilon}{2^{m+1}}$ holds as
for any $t\in(u_i,v_i)$ and letting $ d_i \in(u_i,v_i)$ and $\epsilon > 0,$
$|\alpha(t) - \alpha(d_i)| = |2-2| = 0 < \frac{\epsilon}{2^{m+1}}$
Now let $\epsilon = 1$ which is allowed as above should hold for any $\epsilon >0$.
thus $0< \frac{\epsilon}{2^{m+1}} < 1$.
Note: $0< \frac{\epsilon}{2^{m+1}}*2 = \frac{\epsilon}{2^{m}} <2$
But now $\alpha(v_i) - \alpha(u_i) < \frac{\epsilon}{2^m} < 2$ is not true as:
$\alpha(v_i) - \alpha(u_i) = 3-1 = 2$
thus $\alpha(v_i) - \alpha(u_i) > \frac{\epsilon}{2^m}$ which is a contradiction to $\alpha(v_i) - \alpha(u_i) < \frac{\epsilon}{2^m}$
You are right, the proof does not not work as it is presented. Fortunately it can be easily repaired.
The purpose of the proof is to find $u_i, v_i$ such that $u_i < d_i < v_i$ and $\lvert \alpha(u_i) - \alpha(v_i) \rvert < \varepsilon/2^m$.
Take $u_i, v_i$ as in the proof in your question. Now choose $\bar u_i, \bar v_i$ such that $u_i < \bar u_i < d_i < \bar v_i < v_i$. Then for $t \in [\bar u_i, \bar v_i]$ $$\lvert \alpha(t) - \alpha(d_i) \rvert < \varepsilon/2^{m+1}$$ and we conclude $\lvert \alpha(\bar u_i) - \alpha(\bar v_i) \rvert < \varepsilon/2^m$.
Thus we can replace the original $u_i, v_i$ by $\bar u_i, \bar v_i$.