This is a question from Finite-Dimensional Linear Algebra by Mark S. Gockenbach page 72 (Exercise 2.7.14). I hope to check my proof. Thank you.
Let $V$ be an $n$-dimensional vector space over a field $F$, and suppose $S$ and $T$ are subspaces of $V$ satisfying $S \cap T = \{0\}$. Suppose that $\{s_1,…,s_k \}$ is a basis for $S$, $\{t_1,…,t_k \}$ is a basis for $T$, and $k+l = n$.
Prove that $\{s_1,…,s_k,t_1,…,t_l \}$ is a basis for $V$.
My proof :
Suppose that $\alpha_1 s_1 +…+ \alpha_k s_k + \alpha_{k+1}t_1 +…+ \alpha_{n}t_l = 0$ for some $\{ \alpha_i\}_{i=1}^{n} \subset F$
Then $\alpha_{k+1}t_1 + … + \alpha_{n} t_l = -(\alpha_1 s_1 + … + \alpha_k s_k) \in \text{span} \{ s_1,…,s_k \}= S$.
Also, $\alpha_{k+1}t_1 + … + \alpha_{n} t_l \in \text{span} \{ t_1,…,t_l \} = T $.
By Exercise 2.5.15, $\{s_1,…,s_k,t_1,…,t_l \}$ is a linearly independent set with $n$ vectors.
Since $\text{dim}(V)=n$, any set of $n+1$ vectors is linearly dependent.
If there exists $v \notin \text{span} \{s_1,…,s_k,t_1,…,t_l \} $, then we could add that vector $v$ to the set to obtain a set of $n+1$ linearly independent vectors.
However, since $\text{dim}(V)=n$, any set of $n+1$ vectors is linearly dependent.
This shows that every $v \in V$ belongs to $\text{span} \{s_1,…,s_k,t_1,…,t_l \} $.
Therefore, $\text{span} \{s_1,…,s_k,t_1,…,t_l \} = V$ and is therefore a basis for $V$. $\blacksquare$
Best Answer
Since the vectors are in bases of different subspaces, they are non-zero and linearly independent. Therefore, their union is also a basis for the span of the new set.