Proving that if $v_1+\dots+v_k\in W$ then $v_i\in W$

Let $T$ be a linear operator on a finite dimensional vector space $V$. $W$ is a $T$-invariant subspace of $V$. Let $v_1,\dots,v_k$ be eigenvectors of $T$. How do I prove that if $v_1+\dots+v_k\in W$ then $v_i\in W$?

I tried doing it by induction on $k$, where case $k=1$ is trivial. But I'm lost on how to use the fact that if $v_1,\dots,v_k\in W$ then $v_1,\dots,v_{k+1}\in W$, if $v_{k+1}+w\in W$ (where I simplified $v_1+\dots+v_k=w$).

Edit: The eigenvectors do indeed have distinct eigenvalues

As the other answers point out, your claim is false as stated. However, if the eigenvectors $v_i$ correspond to distinct eigenvalues, then your claim is true.
Suppose that $w := \sum_{i=1}^{k+1} v_k \in W$, where $$\forall 1 \leq i \leq k+1, \quad T(v_k) = \lambda_k v_k$$ with $\lambda_i \neq \lambda_j$ for $i \neq j$. Since $W$ is $T$-invariant, it follows that $T(w) = \sum_{i=1}^{k+1} \lambda_i v_i$ is still in $W$; since $W$ is a subspace of $V$, it then follows that for each $1 \leq j \leq k+1$, $$T(w) - \lambda_jw = \sum_{i \neq j} (\lambda_i - \lambda_j)v_i$$ is still in $W$, so that by the induction hypothesis, $(\lambda_i - \lambda_j)v_i \in W$ for $i \neq j$, and hence, by our extra assumption on the eigenvectors, $v_i \in W$ for $i \neq j$. Doing this for $j=1,2$ is then enough to show that $v_i \in W$ for all $1\leq i \leq k+1$.
This is completely false. Take $T$ to be the identity on $R^2$, with eigenvectors $v_1=(1,0)$ ,$v_2=(0,1)$ and $W=\mbox{span}\{(1,1)\}$, so that $TW=W$. Then clearly $v_1+v_2=(1,1)\in W$ but neither $v_i$ is in $W$.
Let $V= \mathbb{R}^n$, $T = \textrm{Id}$, and $W= \mathbb{R} \left( \begin{matrix} 1 \\ \vdots \\ 1 \end{matrix} \right)$. Then we have that $W$ is $T$-invariant and for the eigenvectors $\{e_i \mid 1 \le i \le n \}$ we have $\sum_{i=1}^n \in W$ but $e_1 \not\in W$.
If we assume that the eigenvectors $v_i$ have distinct eigenvalues, then we can consider the vectors $T^n w \in W$ as $W$ is $T$-invariant, and for $0 \le n \le k-1$ we get $k$ elements in $W$ which are, written w.r.t the "basis" of $\{ v_i \mid 1 \le i \le k\}$ independent, so by cleverly subtracting we find $v_i \in W$ for all $1 \le i \le k$.