I give an answer here using only elementary algebraic geometry.

We know that the Grassmannian $\operatorname{Gr}(2,5)$ is defined by $5$ linearly independent quadrics in $\mathbb{P}^9$ and $V(5)$ can be viewed as defined by $5$ linearly independent quadrics in $\mathbb{P}^6$.

Given a point $p=[1,0,\dots,0]\in X$ for $X\subset \mathbb{P}^6$ defined by $Q_i, 1\leq i\leq 5$, where $Q_i$ are homogeneous of degree $2$.
Write $$Q_i= q_{i,2}(x_1,\dots, x_6)+ x_0 q_{i,1}(x_1,\dots, x_6)$$
with $q_{i,k}$ homogeneous of degree $k$.

I claim that
there there is a line contained in $X$ passing through $p$ if and only if $$q_{i,k}=0, \quad 1\leq i \leq 5, k=1,2 \quad (*)$$ have a common zero in $\mathbb{P}^{5}=\mathbb{P}(k[x_1,\dots x_6])$. To see this, note that given a common zero $[a_1,a_2,\dots,a_6]\in\mathbb{P}^{5}$, the line joining $p$ to $[0, a_1,a_2,\dots,a_6]$ is contained in $X$ and the line joining $[1,0,\dots,0]$ and $[1,-a_1,\dots,-a_n]$ is the same as the line joining $[1,0,\dots,0]$ and $[0, a_1,\dots,a_n]$.

Since $Q_i$ are linearly independent, we see that the linear part $q_{i,1}, 1\leq i\leq 5$ are linearly independent. By the geometric version of Krull’s Principal Ideal Theorem[11.3.2, Vakil’s notes], the dimension of the subvariety in $\mathbb{P}^5$ defined by $(*)$ is $\leq 0$. This means $(*)$ has finitely many common zeros, that is, there are at most finitely many lines passing through a given point.

Now back to $X=V(5)$, as suggested by Enrico in the comment. We take $3$ hyperplanes ''general enough'' to get explicit defining equations of $V(5)$. Consider the Plucker coordinate $w_{ij}$ in $\mathbb{P}^9$, take $w_{12}=w_{34}$, $w_{13}=w_{45}$, $w_{23}=w_{15}$, the defining equations for $V(5)$ are

\begin{cases}
w_{34}^2-w_{45}w_{24}+w_{14}w_{15}=0\\
w_{45}^2-w_{14}w_{35}+w_{15}w_{34}=0\\
w_{15}^2-w_{45}w_{25}+w_{34}w_{35}=0\\
w_{15}w_{45}-w_{24}w_{35}+w_{25}w_{34}=0\\
w_{34}w_{45}-w_{14}w_{25}+w_{15}w_{24}=0
\end{cases}

By the above argument, choose $p$ to be the point with the only nonzero entry $w_{ij}$ for $\{i,j\}\neq \{3,4\},\{4,5\},\{1,5\}$ we can have some lines. For example,
we can find two nonintersecting lines: one joining $[ x_{14}=1,x_{ij}=0]$ and $[x_{12}=1,x_{ij}=0]$, the other joining $[ x_{24}=1,x_{ij}=0]$ and $[ x_{25}=1,x_{ij}=0]$. This has solved the question.

**Remark.**
Suppose we are given the fact that for every point, there is at least one line passing through it, we can show that the subvariety in $\operatorname{Gr}(2,7)$ characterising lines on $V(5)$ has dimension $ 2$. Consider the following incidence correspondence, as we have discussed above, for any $p\in V(5)$ the fibre $\operatorname{pr}_1^{-1} (p)$ has dimension $0$.

By Upper Semicontinuity of Fibre Dimension, we have $\operatorname{dim} \Phi= \operatorname{dim} V(5)=3$. Take a line $l\in V(5)$, $\operatorname{dim}pr_2^{-1}([l])=1$, by Upper Semicontinuity again, $\operatorname{dim}\Phi -\operatorname{dim} pr_2\Phi = 1$, $\operatorname{dim} pr_2\Phi = 2$. We conclude that the subvariety of $\operatorname{Grass}(2,7)$ characterising lines on $V(5)$ has dimension $ 2$.

A stronger result than $\operatorname{dim }pr_2\Phi= 2$ is that the Hilbert scheme of lines on $V(5)$ is isomorphic to $\mathbb{P}^2$ as in Sasha's answer and the reference is [Theorem 1, 2]. And the number of lines passing through a point is $3$ when counted with multiplicity [P.26, 1].

$[1]$
*Takagi, Hiromichi; Zucconi, Francesco*, **Geometries of lines and conics on the quintic del Pezzo 3-fold and its application to varieties of power sums**, Mich. Math. J. 61, No. 1, 19-62 (2012). ZBL1262.14048.

$[2]$*Furushima, Mikio; Nakayama, Noboru*, **The family of lines on the Fano threefold (V_ 5)**, Nagoya Math. J. 116, 111-122 (1989). ZBL0731.14025.