Perturbation of linear independence Given a linearly independent and orthonormal set $\lbrace u_1,\ldots,u_n\rbrace \in \mathbb{R}^d$.  Exist $0<\varepsilon<1$ such that $B(u_i,\varepsilon)\cap B(u_j,\varepsilon)= \emptyset$ for $i\neq j$ where $1\leq i,j \leq n$. 
Consider $\{v_j \}$ such that $\Vert v_j- u_j\Vert< \varepsilon$ for all $1\leq j \leq n$.  I wonder if the set $\lbrace v_1,\ldots,v_n\rbrace$ is linearly independent ? 
My progress: If $\sum_{j=1}^n \alpha_j v_j=0$ then $v_j=\sum_{k=1}^n \beta_k u_k$ therefore $\sum_{j=1}^n  \sum_{k=1}^n \alpha_j\beta_k u_k=0$ then obtain using linear independence $\sum_{j=1}^n \alpha_j=0$ .... (1) 
Also $\sum_{j=1}^n \alpha_j (v_j - u_j)=-\sum_{j=1}^n \alpha_j u_j$ then $\Vert \sum_{j=1}^n \alpha_j u_j\Vert<\varepsilon \sum_{j=1}^n |\alpha_j |$ .... (2) 
Would greatly appreciate any suggestions to complete the test

Your inequality 
$$\Big\Vert \sum_{j=1}^n \alpha_j u_j\Big\Vert<\varepsilon \sum_{j=1}^n |\alpha_j |\tag2$$
is indeed useful here. By orthonormality, the square of the left side is $\sum_{j=1}^n \alpha_j^2$. The square of the right side can be estimated using the Cauchy-Schwatz inequality
$$\Big(\varepsilon \sum_{j=1}^n |\alpha_j| \Big)^2 \le \varepsilon^2 n \sum_{j=1}^n \alpha_j^2$$
Hence $1<\varepsilon^2n$. In other words, we reach a contradiction provided that $\varepsilon\le 1/\sqrt{n}$.

And this is the best you can get. The assumption $B(u_i,\varepsilon)\cap B(u_j,\varepsilon)= \emptyset$, which allows $\varepsilon$ as large as $1/\sqrt{2}$, does not suffice. For example, in three dimensions we can perturb the standard basis vectors $$(1,0,0), (0,1,0), (0,0,1)$$
to $$(2/3,-1/3,-1/3), (-1/3, 2/3,-1/3), (-1/3,-1/3,2/3)$$
which are linearly dependent. None of the vectors moved by more than $1/\sqrt{3}$. The example generalizes to higher dimensions: just project the standard basis onto the hyperplane $x_1+\dots+x_n=0$. 
