Let's show one possible approach on an example. Suppose we are given a set of vectors $(1,-2,-2,2),(2,0,-1,1),(3,0,-4,-4)$ in $\mathbb R^4$. We want to know whether their span $V_A=[(1,-2,-2,2),(2,0,-1,1),(3,0,-4,-4)]$ contains the span $V_B=[(1,0,1,1),(1,1,0,0)]$ of the vectors $(1,0,1,1)$ and $(1,1,0,0)$. This is equivalent to finding out whether $(1,0,1,1)\in V_A$ and $(1,1,0,0)\in V_A$.
First we put the vectors from the set $A$ as a rows into a matrix and we use elementary row operations to get a row echelon form.
$ \begin{pmatrix} 1 & -2 & -2 & 2 \\ 2 & 0 & -1 & -1 \\ 3 & 0 & -4 & -4 \end{pmatrix} \overset{(1)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 4 & 3 & -5 \\ 0 & 0 & -\frac52 & -\frac52 \end{pmatrix} \overset{(2)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 4 & 3 & -5 \\ 0 & 0 & 1 & 1 \end{pmatrix} \overset{(3)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 4 & 0 & -8 \\ 0 & 0 & 1 & 1 \end{pmatrix} \overset{(4)}{\sim} \begin{pmatrix} 1 & -2 & -2 & 2 \\ 0 & 1 & 0 & -2 \\ 0 & 0 & 1 & 1 \end{pmatrix} \overset{(5)}{\sim} \begin{pmatrix} \boxed1 & 0 & 0 & 0 \\ 0 & \boxed1 & 0 & -2 \\ 0 & 0 & \boxed1 & 1 \end{pmatrix} $
Now we know that $V_A=[(1,0,0,0),(0,1,0,-2),(0,0,1,1)]$ since row operations do not change the row space.
But since we know have simpler vectors generating $V_A$, it is much easier to find out whether some given vector belongs to $V_A$.
If some vector $(a_1,a_2,a_3,a_4)$ belongs to $V_A$, it must fulfill $(a_1,a_2,a_3,a_4)=a_1(1,0,0,0)+a_2(0,1,0,-2)+a_3(0,0,1,1)$. (It suffices to look at the position of the leading coefficients of the row echelon form; in this case the first three positions. The leading coefficients are marked by boxes in the row echelon matrix.)
Now we want test the vectors $(1,0,1,1)$ and $(1,1,0,0)$.
For the first one we get $1\times(1,0,0,0)+1\times(0,0,1,1)=(1,0,1,1)$, so this vectors belongs to $V_A$.
For the second one we get $1\times(1,0,0,0)+1\times(0,1,0,-2)=(1,1,0,-2)\ne(1,1,0,0)$. Therefore $(1,1,0,0)\notin V_A$ and, consequently, $V_B\not\subseteq V_A$.