If $y_1,...,y_l$ are vectors in vector space V and $x_i\in \text{ span }\{y_1,...y_l\}\ \forall i=1,...,k$, how to disprove that span$\{x_1,...,x_k\}=\text{ span }\{y_1,...,y_1\}$.
In my perspective, either $\text{ span }\{x_1,...,x_k\}\subsetneq\text{ span }\{y_1,...,y_1\}$ or $\text{ span }\{x_1,...,x_k\}\supsetneq \text{ span }\{y_1,...,y_1\}$ or both.
Since $x_i\in \text{ span }\{y_1,...y_l\}\ \forall i=1,...,k$, $\text{ span }\{x_1,...,x_k\}=a_1y_1+...+a_ly_l$, meaning any vectors in $\text{ span }(X)$ can be linearly combined by vectors in $\text{ span }(Y)$. Therefore $\text{ span }\{x_1,...,x_k\}\subseteq \text{ span }\{y_1,...,y_1\}$.
It is possible that vectors $\text{ span }(X)$ are $0$ and therefore they cannot linearly combine to for all possible vectors in $\text{ span }(Y)$, and thus $\text{ span }\{x_1,...,x_k\}\supsetneq \text{ span }\{y_1,...,y_1\}$. So $\text{ span }\{x_1,...,x_k\}\ne \text{ span }\{y_1,...,y_1\}$
However, I sensed my proof redundant and incorrect. Could anyone fix it?
To disprove a statement all you need is one counterexample.
As you observed you have for sure $\mbox{ span } \{ x_1,.., x_k\} \subset \mbox{ span } \{ y_1,.., y_l\} $. So this part cannot fail.
This means that you must come up with a single example such that $\mbox{ span } \{ y_1,.., y_l\}$ is not inside $\mbox{ span } \{ x_1,.., x_k\}$. And as you said picking $x_i=0$ seems like a good idea.
So build your example: How many $y's$ do you think you need, and what can they be? How many $x'$s, and what do you want them to be?