It is mostly correct, but incomplete.
You don't state what your $v$ is in the proof. You need to pick $v$ to be an element of $\mathbb R^n$ which is not in the span of $X$. (Thus, you are using one condition.)
You should also state why $\exists \alpha_1,\dots,\alpha_k.$
If $X\cup\{v\}$ is linearly dependent, then $\exists \beta_0,\dots,\beta_k$ such that $$0=\beta_0 v + \beta_1 x_1+\dots+\beta_k x_k$$ with the $\beta_i$ not all zero.
But if $\beta_0=0$ then $X$ would be linearly dependent. So $\beta_0\neq 0$ and from there you can conclude that $\alpha_i=\frac{-\beta_i}{\beta_0}$ gives you:
$$v=\alpha_1 x_1+\cdots+\alpha_kx_k$$
showing $v\in \operatorname{Span}(X),$ which is a contradiction.
Indeed, this could be stated as a general lemma:
Lemma: If $V$ is a vector space and $X\subset V$,
then, for any $v\in V\setminus X$, $X\cup\{v\}$ is linearly independent if and only if $X$ is linearly independent and $v\notin\operatorname{Span}(X).$
This lemma is exactly why you can find $\alpha_1,\dots,\alpha_k.$ Since $X$ is linearly independent, we have, by this lemma, if $X\cup \{v\}$ is linearly dependent, then $v\in\operatorname{Span}(X).$
Aside: This lemma assumes the definition that $\operatorname{Span}(\emptyset)=\{0\}.$
The question has two conditions:
- $X$ is linearly independent
- $X$ does not span $\mathbb R^n$
Without $(2),$ you cannot find $v.$
Without $(1),$ you cannot prove $X\cup\{v\}$ is linearly independent.
Your proof does not mention where you are using either condition.
We can prove this by contradiction without the use of a basis.
First notice that as said in the comments, the hypothesis is that for any linearly independent subsets $S\subseteq U,\,V\subseteq V$, $S\cup T$ is a linearly independent set.
Suppose $x\ne0$ is an element in $U\cap V$. Then $\left\{x\right\}\subseteq U$ is a linearly independent subset. Let $S=\left\{x\right\}$. Also, $\left\{2x\right\}\subseteq V$ is a linearly independent subset. Let $T=\left\{2x\right\}$. Then by the hypothesis, $S\cup T=\left\{x,2x\right\}$ should be a linearly independent set. But this is false: $-2\cdot x+(2x)=0$. Therefore $U\cap V=\left\{0\right\}$.
You may wonder if the conclusion still holds when the hypothesis is changed so that for some specific $S=\left\{u_1,\ldots,u_j\right\}$ and $T=\left\{v_1,\ldots,v_k\right\}$, $S\cup T$ is linearly independent. This is clearly false. For example, take $n=2,\,U=V=\mathbb R^2,\,S=\left\{(1,0)\right\},\,T=\left\{(0,1)\right\}$. Then $S$ and $T$ are linearly independent, and so is $S\cup T$, but $U\cap V=\mathbb R^2\ne\left\{0\right\}$.
Hope this helps.
Best Answer
The proof idea is as follows:
Since $L$ does not span $V$ it is true that $$V \backslash \text{span}(L) \neq \emptyset$$ That implies that there exists an element in $V$ that is not in the span of $L$ and thus cannot be written as a linear combination of the other elements in the span of $L$ which means there exists an element, call it $v$, in $V$ such that $L \cup \{v\}$ is linearly independent.