Let $\left\{ \omega_k \right\}_{k \geq 0}$ a sequence of not decreasing value such that
$$ \omega_k \leq \sum_{j\geq k+1} \omega_j $$
Let $t \in A = \left[-\sum_{j\geq 0}\omega_j, \sum_{j\geq 0}\omega_j \right]$ and $t_k$ defined as
\begin{equation} t_{k} = \left\{ \begin{array}{ll} 0 & k = 0 \\ t_{k-1} + d_{k-1}(t)\omega_{k-1} & k \geq 1 \end{array} \right. \end{equation} where $$ d_k(t) = \left\{ \begin{array}{rl} 1 & t \geq t_k \\ -1 & t < t_k \end{array} \right. $$ It can be proved by induction that $$ -\sum_{j\geq k}\omega_k \leq t - t_k \leq \sum_{j\geq k} \omega_k. $$
For all natural $k \geq 0$ let $f_{k}(t)$ defined in $A$ as \begin{equation} f_k(t) = t - t_k \end{equation} And let's consider the family of functions \begin{equation} \mathcal{F} = \left\{ f_k \right\}_{k\geq0} \end{equation} I want prove/disprove that for all $f_k \in \mathcal{F}$ $f_k$ has a maximum/minimum value, given the property of the difference $f_k(t) = t - t_k$ we know it is bounded, and by the least upper bound we have that $f_k$ has supremum/infimum. By induction I was trying to prove that $f_k$ has such property. For $k = 0$ we have $$ f_0(t) = t $$ which is the identity function, but defined in a bounded interval, so it is bounded. Assuming until $k$ the property holds I was studying the case $k+1$, we have that
$$ f_{k+1}(t) = f_{k}(t) - d_k(t) \omega_k $$
But I'm stuck here, I was trying to use $\hat{t}$ as argmax/argmin of $f_k$, however such value doesn't guarantee that $f_{k+1}$ achieve the max/min for the same value.
Update:
The other clue I'm having right now is the following. We clearly have
$$ f_k(t) = t - \sum_{0 \leq j \leq k-1} d_j(t) \omega_j = t - <D_{k-1}(t),\Omega_{k-1}> $$ where $$ \begin{array}{l} D_{k-1}(t) = \left(d_0(t),\ldots,d_{k-1}(t) \right) \\ \Omega_{k-1} = \left(\omega_0,\ldots,\omega_{k-1} \right) \end{array} $$
For all $k$ I can then write $A$ as finite union of disjoint set $\left\{A_j\right\}_{j \in J}$ where each $A_j$ is defined as
$$ A_j = \left\{ t \in A : D_{k-1}(t) = \text{constant}_j \right\} $$
So we would have that $$ \max_{t \in A} \left\{ f_k(t) \right\} = \max_{j \in J} \left\{ \max_{t \in A_j} \left\{ f_k(t) \right\} \right\} $$
Given that for all $j \in J$ the restriction $f_k |_{A_j}$ is continuous and defined in a bounded interval we have that that $f_k$ has maximum, $J$ is finite, so the max of all the max exists as well.
Not too much rigorous, but that's probably the way to go.