Proofing dependency in linear algebra involving span

Assume

1. $x_1, x_2, x_3, x_4, x_5 \in \mathbb{R}^m$

2. $(x_1, x_2, x_3)$ is linearly independent

3. $(x_4, x_5)$ is linearly independent.

4. Span$(x_1, x_2, x_3)$ ∩ Span$(x_4, x_5)$ $\neq$ {0} (this means that there are non-zero vectors common to Span$(x_1, x_2, x_3)$ and Span$(x_4, x_5)$. Prove that $(x_1, x_2, x_3, x_4, x_5)$ is linearly dependent.

I have difficulty in proofing this, can someone please show me how to do this?

Suppose the set $\vec x$ is linearly independent. The 4th proposition, implies that there is a $y \neq 0$ such that $$a x_1 + b x_2 + c x_3 = y = d x_4 + e x_5,$$ with $abc \neq 0$ and $de \neq 0$. This implies that $$a x_1 + b x_2 + c x_3 - d x_4 - e x_5 = 0.$$ Which proves that the set is linearly dependent

By 4 we can write

$a_1 x_1 + a_2 x_2 + a_3 x_3 = b_1 x_4 + b_2 x_5$

for some $a_i$'s, $b_j$'s. Then there exists some nonzero $a_i$'s and $b_j$'s such that

$a_1 x_1 + a_2 x_2 + a_3 x_3 - b_1 x_4 - b_2 x_5 = 0$

which proves they are linearly dependent.

Since $\mathrm{Span}\{x_1,x_2,x_3\}\cap \mathrm{Span}\{x_4,x_5\}=\{0\},$ then this is to say that for any scalars $a,b,$ if $v=ax_4+bx_5,$ then $v\notin \mathrm{Span}\{x_1,x_2,x_3\}.$ In particular, if $a=1, b=0,$ then $x_4$ is not a linear combination of $x_1, x_2, x_3,$ which means that $\{x_1,x_2,x_3,x_4\}$ is linearly independent. Since $\{x_4,x_5\}$ is linearly independent, and if $a=0, b=1,$ then $x_5$ is not a linear combination of $x_1, x_2, x_3,$ then $\{x_1,x_2, x_3, x_4,x_5\}$ must be linearly independent.