$T = \{v_1+\epsilon,v_2+\epsilon,\cdots,v_n+\epsilon \}$ is linearly independent. Suppose we have a set of linearly independent vectors say $S = \{v_1,v_2,\cdots,v_n \}$, then we can find some $\epsilon > 0$ such that $T = \{v_1+\epsilon,v_2+\epsilon,\cdots,v_n+\epsilon \}$ is still linearly independent.
Note that by $v + \epsilon$ we mean we are adding $\epsilon$ to each of the coordinates.
This result is probably known as the stability of linearly independent vectors. I am not sure. Can someone provide some reference which includes the proof?

Let's denote by $u$ the vector whose all coordinates are $1$. We will prove that if $(v_1, ..., v_n)$ are linearly independent, there is at most one value of $\varepsilon$ such that $(v_1 + \varepsilon u, ..., v_n + \varepsilon u)$ are not linearly independent. Of course, this is a stronger result that the one you want to prove.
Let's suppose that there exists $\varepsilon \neq \varepsilon'$ such that both families $(v_1 + \varepsilon u, ..., v_n + \varepsilon u)$ and $(v_1 + \varepsilon' u, ..., v_n + \varepsilon' u)$ are not linearly independent. Then, there exists $(\lambda_1, ...,\lambda_n) \neq (0, ..., 0)$ and $(\mu_1, ..., \mu_n) \neq (
0, ..., 0)$ such that
$$\lambda_1(v_1 + \varepsilon u) + ... + \lambda_n(v_n + \varepsilon u) = 0$$
$$\mu_1(v_1 + \varepsilon' u) + ... + \mu_n(v_n + \varepsilon' u) = 0$$
i.e.
$$\lambda_1v_1 + ... + \lambda_nv_n= -(\lambda_1 + ... + \lambda_n)\varepsilon u $$
$$\mu_1v_1 + ... + \mu_nv_n= -(\mu_1 + ... + \mu_n)\varepsilon'u $$
Notice that $\lambda_1 + ... + \lambda_n \neq 0$, and $\varepsilon \neq 0$, because $(v_1, ..., v_n)$ are linearly independent. Then you deduce that
$$\mu_1v_1 + ... + \mu_nv_n = \frac{(\mu_1 + ... + \mu_n)\varepsilon'}{(\lambda_1 + ... + \lambda_n)\varepsilon}\left(\lambda_1v_1 + ... + \lambda_nv_n \right)$$
Because the $(v_1, ..., v_n)$ are linearly independent, you deduce that for all $k=1, ..., n$,
$$\mu_k = \frac{(\mu_1 + ... + \mu_n)\varepsilon'}{(\lambda_1 + ... + \lambda_n)\varepsilon} \lambda_k$$
and summing for $k=1, ..., n$, you get
$$\mu_1 + ... + \mu_n = \frac{(\mu_1 + ... + \mu_n)\varepsilon'}{\varepsilon}$$
But $\mu_1 + ... + \mu_n$ cannot be $0$,  because $(v_1, ..., v_n)$ are linearly independent, so you deduce $\varepsilon = \varepsilon'$, which is absurd.
