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 $\geq 2$. 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 $0$.
[![enter image description here][1]][1]
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, $\operatorname{dim}\Phi -\operatorname{dim} pr_2\Phi \leq 1$, $\operatorname{dim} pr_2\Phi \geq 2$. We conclude that the subvariety of $\operatorname{Grass}(2,7)$ characterising lines on $V(5)$ has dimension $\geq 2$.
A stronger result than $\operatorname{dim }pr_2\Phi\geq 2$ 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. [1]: https://i.sstatic.net/M6cjS.png
