Let $\{f_t:t\geqslant 0\}$ be a sequence of measurable functions on a countable space $\mathcal X$ and $\pi$ a measure on $\mathcal X$ with $\sum_{y\in \mathcal X}\pi(y)=M<\infty$ Suppose that $\lim_{t\to\infty}f_t(y)=0$ for all $y\in\mathcal X$ and there exists $B<\infty$ such that $|f_t(y)|\leqslant B$ for all $t,y$. Then
$$\lim_{t\to\infty}\sum_{y\in\mathcal X}\pi(y)f_t(y) = 0. $$
My argument is that since $\{f_t\}$ is uniformly bounded, we may pick $B<\infty$ such that $|f_t(y)|\leqslant B$ for all $t,y$, from which for any $\varepsilon>0$ we may choose $T$ large enough that $f_t(y) <\frac\varepsilon{MB}$ for $t\geqslant T$. Then for $t\geqslant T$ we have $$ \sum_{y\in\mathcal X}\pi(y)f_t(y) \leqslant MB\cdot\frac\varepsilon{MB} = \varepsilon, $$ from which it follows that the limit is zero. Is this correct?
Your proof is not valid since $T$ may depend on $y$. This result is immediate consequence of DCT.
For a proof without DCT note that the tail sums of the series $\sum_y \pi (y)$ tend to $0$. Hence there is a finite set $F=\{y_1,y_2,...,y_N\}$ such that $\sum_{y \notin F}\pi (y) <\epsilon$. Now $\sum f(y)\pi (y) = \sum_{y \notin F} f(y)\pi (y)+\sum f(y_i)\pi (y_i)$. Use the bound for $f$ in the first term and let $t \to \infty$ in the second term.