No. Look at the plane $\mathbb R^2$ with the usual basis vectors $(1,0)$ and $(0,1)$ and the subspace $A = \{(x,x) : x \in \mathbb R\}$.
Well,think about what it means to have a basis for a vector space V over a field F. A basis B is a set of vectors in V for which for every vector v in V, there exists S= ${v_1,v_2,.....v_n}\subseteq B$ and ${a_1,a_2,...,a_n}\in F$ such that $\sum_{i=1}^n a_iv_i = v$ and $\sum_{i=1}^n a_iv_i = 0$ iff for every $i$, $a_i=0$. So let's check. Let U be a subspace of V and consider the quotient space V\U. Consider $S'= S + U\subseteq V/U $ and let's see if this is a basis for V/U. Let's see if it spans V/U. For every $u\in V/U$, u = v + w where v is an arbitrary vector in V where $u-v=w\in U$. Since B is a basis for V, there exists ${v_1,v_2,.....v_i}\in B$ and ${a_1,a_2,...,a_i}\in F$ where $1\leq i \leq n$ such that u = $\sum_{j=1}^i a_jv_j$. Also, since B is a basis for V and U is a subspace of V, there exists ${v_1,v_2,.....v_k}\in B$ and ${b_1,b_2,...,b_k}\in F$ such that for every $w\in U$, w = $\sum_{l=1}^k b_lv_l$ where $1\leq k \leq n$. But this means u = v + w = $\sum_{j=1}^i a_jv_j$ + $\sum_{l=1}^k b_lv_l$ = $\sum_{m=1}^{j+l}(a_j + b_l)v_m$ where $1\leq j+l \leq n$ . But this means S' spans V/U. Since for every m where $1\leq m\leq n$, $v_m\in B$, then $\sum_{i=1}^m a_iv_i = 0$ iff for every $i\leq m$, $a_i=0$. But that means S' is a linearly independent set of vectors in V and that means S' is a basis for V/U. Q.E.D.
The notation of my proof in the indices may be a little sloppy. I'll go over it later, but the basic logic is correct.
Best Answer
Certainly not: For any field $\mathbb{F}$, consider the basis $((1, 0), (0, 1))$ of $\mathbb{F}^2$. Neither $(1, 0)$ nor $(0, 1)$ spans the subspace spanned by $(1, 1)$.