We don't necessarily know that $\{v_i:1\le i\le k\}$ is a basis of $V$ since we don't know the dimension of $V$, but we are given that $\{v_i\}$ are orthonormal; that is $\left=0$ when $i\not=j$ and $\left=1$.
Consider the vector
$$
x^\perp=x-\sum_{i=1}^k\leftv_i\tag{1}
$$
$x^\perp$ is perpendicular to $\{v_i\}$:
$$
\begin{align}
\left
&=\leftv_i,v_j\right>\\
&=\left-\left\left\\
&=0\tag{2}
\end{align}
$$
Therefore, $x^\perp$ is perpendicular to $x-x^\perp=\sum\limits_{i=1}^k\leftv_i$.
Next consider
$$
\begin{align}
\left
&=\left<\sum_{i=1}^k\leftv_i,\sum_{j=1}^k\leftv_j\right>\\
&=\sum_{i=1}^k\left\left\left\\
&=\sum_{i=1}^k\left\left\tag{3}
\end{align}
$$
Note that since $x^\perp$ is perpendicular to $x-x^\perp$,
$$
\|x-x^\perp\|^2+\|x^\perp\|^2=\|x\|^2\tag{4}
$$
which implies that $\|x-x^\perp\|\le\|x\|$.
Now $(3)$, Cauchy-Schwarz, and $(4)$ yield
$$
\begin{align}
\left|\sum_{i=1}^k\left\left\right|
&=\left|\left\right|\\
&\le\|x-x^\perp\|\|y-y^\perp\|\\
&\le\|x\|\|y\|\tag{5}
\end{align}
$$
To finish off the proof (thanks to cardinal), consider the vector
$$
x^+=\sum_{i=1}^k\left|\left\right|v_i\tag{6}
$$
Note that $\|x^+\|^2=\sum\limits_{i=1}^k\left^2=\|x-x^\perp\|^2$.
Plugging $x^+$ and $y^+$ into $(5)$ gives
$$
\begin{align}
\sum_{i=1}^k\left|\left\left\right|
&=\left|\sum_{i=1}^k\left\left\right|\\
&\le\|x^+\|\|y^+\|\\
&\le\|x\|\|y\|\tag{7}
\end{align}
$$