Prove that if $e_1, \dots e_n$ is an orthonormal basis of $ V$ and $v_1, \dots , v_n$ are vectors in $ V$ such that $$\|e_j - v_j\| < \frac{1}{\sqrt{n}}$$ for each j, then $v_1, \dots v_n$ is a basis of $V$.
I have fiddled around with this for a while now unfortunately nothing came close, so I'll just say I have no idea how to do this. Help :(?
Although not explicitly said in the question, we are assuming that $V$ is an inner product space.
Suppose that $v_1,\ldots,v_n$ is linearly dependent. Then there exist constant, not all zero, $c_1,\ldots, c_n$ with $c_1v_1+\cdots+c_nv_n=0$. Then \begin{align} \|c_1e_1+\cdots+c_ne_n\|&=\|c_1(e_1-v_1)+\cdots+c_n(e_n-v_n)\|\\ \ \\ &\leq\sum_{j=1}^n|c_j|\,\|e_j-v_j\|\leq\sum_{j=1}^n\frac{|c_j|}{\sqrt n}. \end{align} We also know that $$ \|c_1e_1+\cdots+c_ne_n\|^2=\sum_{j=1}^n|c_j|^2, $$ So we have shown that $$ \left(\sum_{j=1}^n|c_j|^2\right)^{1/2}\leq\sum_{j=1}^n\frac{|c_j|}{\sqrt n}. $$ This is the reverse of the Cauchy-Schwarz inequality, so we conclude that $$ \left(\sum_{j=1}^n|c_j|^2\right)^{1/2}=\sum_{j=1}^n\frac{|c_j|}{\sqrt n}, $$ and equality in Cauchy-Schwarz implies that there is a constant $d$ such that $|c_j|=d/\sqrt n$ for all $j$. So $c_j=\lambda_j\,d/\sqrt n$ with $|\lambda_j|=1$, and $$ 0=c_1v_1+\cdots+c_nv_n=\frac d{\sqrt n}\,(\lambda_1v_1+\cdots+\lambda_n v_n), $$ implying that $\lambda_1v_1+\cdots+\lambda_n v_n=0$ (this works because $d\ne0$, which follows from the fact that the $c_j$ cannot all be zero). Then \begin{align} \sqrt n&=\|\lambda_1e_1+\cdots+\lambda_n e_n\|=\|\lambda_1(e_1-v_1)+\cdots+\lambda_n(e_n-v_n)\|\\ \ \\ &\leq\sum_{j=1}^n\|e_j-v_j\|<\sum_{j=1}^n\frac1{\sqrt n}=\sqrt n, \end{align} a contradiction. This shows that it is impossible that $v_1,\dots,v_n$ are linearly dependent.
WLOG suppose $V=\mathbb C^n$ with standard inner product and $(e_1,\ldots,e_n)$ is the standard basis. Let $A$ be the $n$-by-$n$ matrix with columns $v_1,\ldots,v_n$. Then the Frobenius norm of $I-A$ is the square root of $\sum\limits_{k=1}^n\|e_k-v_k\|^2<1$. Consequently, $A$ is invertible (with inverse $\sum\limits_{k=0}^\infty (I-A)^k$), and therefore it has linearly independent columns.
