Assume that we have a sequence $\{\mathbf{x}_{n}\}_{n=1}^{\infty}$ of non-negative vectors in $\ell_{1}$, i.e. $x_{n,i}\geq 0$ for all $n = 1,2 \dots $ and $i= 1,2 \dots $. Next, assume that for any $i$ we have $$ x_{n,i} \to a_{i} $$ and $||\textbf{x}_{n}||_{1} =||\textbf{a}||_{1} = 1$. Then, as was shown in convergence in $\ell_{1}$, the whole sequence $\textbf{x}_{n}$ converges to $\textbf{a}$ in $\ell_{1}$.
Next, consider non-negative sequence $\textbf{y}_{n}$, such that $y_{n,i} \leq x_{n,i}$ for all $n = 1,2 \dots $ and $i= 1,2 \dots $. Also, assume that $\textbf{y}_{n}$ converges to $\textbf{a}$ pointwise. The proposition: sequence $\textbf{y}_{n}$ converges to $\textbf{a}$ in $\ell_{1}$.
Attempt:
Let's fix some $\delta>0$. Then there exists $M$ such that $$ \sum_{i\leq M}a_{i} \geq 1 - \frac{\delta}{2}. $$
Since both $\mathbf{x}_{n}$ and $\mathbf{y}_{n}$ converge to $\mathbf{a}$ pointwise, then there exist $n_{1}$ and $n_{2}$ such that for all $n\geq n_{1}$ we have $$ \sup_{i\leq M}|x_{n,i} - a_{i}| \leq \frac{\delta}{2M}. $$ and for all $n\geq n_{2}$ we have $$ \sup_{i\leq M}|y_{n,i} - a_{i}| \leq \frac{\delta}{2M}. $$
It follows that $\sum_{i\leq M}x_{n,i} \geq 1 - \delta$ for all $n\geq n_{1}$, and $\sum_{i\leq M}|y_{n,i} - a_{i}| \leq \frac{\delta}{2}$ for all $n\geq n_{2}$.
Next $$ \sum_{i}|y_{n,i} - a_{i}| = \sum_{i\leq M}|y_{n,i} - a_{i}| + \sum_{i>M}|y_{n,i} - a_{i}| \leq \sum_{i\leq M}|y_{n,i} - a_{i}| + \sum_{i>M} y_{n,i} + \sum_{i>M}a_{i}. $$ Next, since $0<y_{n,i} \leq x_{n,i}$, then for all $n>\max(n_{1},n_{2})$ we have $$ \sum_{i\leq M}|y_{n,i} - a_{i}| + \sum_{i>M} y_{n,i} + \sum_{i>M}a_{i} \leq \frac{\delta}{2} + \delta + \frac{\delta}{2} = 2 \delta $$
Therefore, we have shown that for any $\delta>0$ there exists $n$, such that $$ ||\textbf{y}_{n} - \textbf{a}|| \leq 2 \delta, $$ which proves the statement.
The proof is correct in my opinion.
Maybe as an intermediate step it would be nice to explicitly mention $\sum_{i>M} x_{n,i} \leq\delta$, so that it will be easier to see why $\sum_{i>M} y_{n,i} \leq\delta$ holds.