Let $B_{\lambda_i}$ have $n_i$ elements: $v_{i,1}, v_{i,2}, \dots, v_{i,n_i}$. Let's try your induction; suppose there is some nonzero $u \in \left(B_{\lambda_1}\cup\dots\cup B_{\lambda_n}\right)\cap B_{\lambda_{n+1}}$. Then there is a linear combination
$$\tag{1}\label{eq1}\sum_{i=1}^n\sum_{j=1}^{n_i}c_{i,j}v_{i,j}=u$$
By applying $\Phi$ to both sides we obtain that
$$\tag{2}\label{eq2}\sum_{i=1}^n \sum_{j=1}^{n_i}\big(\lambda_i c_{i,j}\big)v_{i,j}=\lambda_{n+1}u$$
If $\lambda_{n+1}=0$, then we have a linear combination on $B_{\lambda_1}\cup\dots\cup B_{\lambda_n}$ that equals $0$.
Moreover, since the $\lambda_i$'s are all distinct, they must all be nonzero for $i<n+1$.
It then follows from the induction hypothesis that the $c_{i,j}$ must be all $0$, contradicting the assumption that $u\neq0$.
Now, if $\lambda_{n+1}\neq 0$, then we may write
$$u=\sum_{i=1}^n \sum_{j=1}^{n_i}\left(\frac{\lambda_i}{\lambda_{n+1}} c_{i,j}\right)v_{i,j}.$$
Then, combining equations $\eqref{eq1}$ and $\eqref{eq2}$ we obtain
$$0=u-u=\sum_{i=1}^n \sum_{j=1}^{n_i}\left(\frac{\lambda_i}{\lambda_{n+1}}-1\right)c_{i,j}\,v_{i,j},$$
so that by the induction hypothesis $\left(\frac{\lambda_i}{\lambda_{n+1}}-1\right)c_{i,j}=0$ for all $i,j$.
Since the $\lambda$'s are distinct, we have that $\frac{\lambda_i}{\lambda_{n+1}} \neq 1$ so it must be that the $c_{i,j}$ are all $0$, again in contradiction with the assumption that $u\neq0$.
We see that whatever the case, there ca be no nonzero $u\in \left(B_{\lambda_1}\cup\dots\cup B_{\lambda_n}\right)\cap B_{\lambda_{n+1}}$, so the inductive step is complete.
Suppose $X \cup Y$ is not linearly independent.
So, there exists $x_1, ... ,x_n \in X $ and $y_1, ..., y_n \in Y$ s.t. $\lambda_1 x_1 + ... + \lambda_n x_n + \alpha_1 y_1 +...+ \alpha_n y_n = 0$.
$\lambda_1 x_1 + ... + \lambda_n x_n = -(\alpha_1 y_1 +...+ \alpha_n y_)$
$(\lambda_1 x_1 + ... + \lambda_n x_n) \in$ $span(X)$, which is contained in $U$, since U is a subspace.
$-(\alpha_1 y_1 +...+ \alpha_n y_) \in$ $span(Y)$ which is in $V$, similarly.
But $span(U \cap V)=\{0\}$, so the only solution to $\lambda_1 x_1 + ... + \lambda_n x_n = -(\alpha_1 y_1 +...+ \alpha_n y_)$ is if all of the scalers, $\lambda$ and $\alpha$ are $=0$
thus shows linear independence in $X \cup Y$
Best Answer
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.