The union of two linearly independent subsets

This is a question from Finite-Dimensional Linear Algebra by Mark S. Gockenbach page 57 (exercise 2.5.15). I hope to check my proof.

Let $V$ be a vector space over a field $F$, and suppose $S$ and $T$ are subspaces of $V$ satisfying $S \cap T = \{0\}$. Suppose $\{s_1,s_2, ..., s_k\} \subset S$ and $\{t_1,t_2, ..., t_l\} \subset T$ are both linearly independent sets.

Prove that $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$.

My proof)

Suppose that $\alpha_1 s_1 +...+ \alpha_k s_k + \alpha_{k+1}t_1 +...+ \alpha_{k+l}t_l = 0$ for some $\{ \alpha_i\}_{i=1}^{k+l} \subset F$

Then $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = -(\alpha_1 s_1 + ... + \alpha_k s_k) \in S$

Since $\{ t_1,...t_l \} \subset T$ and $T$ is a subspace of $V$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_1 \in T $.

Since $S \cap T = \{0 \}$, $\alpha_{k+1}t_1 + ... + \alpha_{k+l} t_l = 0$.

Since $\{t_1, ... , t_l \}$ is a linearly independent set, $\alpha_{k+1} = ... = \alpha_{k+l} = 0$.

By doing similarly, we can show that $\alpha_1 = ... = \alpha_k = 0$.

Therefore, $\{s_1,s_2, ..., s_k,t_1,t_2, ..., t_l \}$ is a linearly independent subset of $V$. $\blacksquare$

It's good, have faith in yourself. :)

Below is an indirect proof,

Suppose on the contrary that $W=\{s_1,s_2,\ldots,s_k,t_1,t_2,\ldots,t_l\}$ is a linearly dependent subset of $V$.

Then, there is a non trivial linear combination of vectors in $W$ whose sum is $\mathbf{0}$.

That is, there exists scalars $a_1,a_2,\ldots,a_k,b_1,b_2,\ldots b_l\in F$, not all zero, such that

$$a_1s_1+a_2s_2+\ldots+a_ks_k+b_1t_1+b_2t_2+\ldots+b_lt_l=\mathbf{0}$$

Case 1: If $a_1=a_2=\ldots=a_k=0$ then $t_1,t_2,\ldots,t_l$ are linearly dependent.

Case 2: If $b_1=b_2=\ldots=b_l=0$ then $s_1,s_2,\ldots,s_k$ are linearly dependent.

If not Case 1 and not Case 2, then

$$a_1s_1+a_2s_2+\ldots+a_ks_k=-b_1t_1-b_2t_2-\ldots-b_lt_l$$

implies that

$$s'=t'$$

for some $s'\in S-\{\mathbf{0}\}$, $t'\in T-\{\mathbf{0}\}$.

In each of the 3 cases, we have a contradiction.