I agree with you, the sentence looks a bit strange to me too. However, I believe that the conclusion that $E_1$ is closed is true. Here is how I would prove it.

Let $(x_j \mid j \geq 1)$ be a sequence of points in $E_1$, converging to some point $x \in \Gamma$. Write $\gamma_j$ for the minimizing geodesic from $q$ to $x_j$, each of which has a non-zero Jacobi field by assumption.

We can extract a subsequence in order to guarantee that $\gamma_j \to \gamma$, a minimizing geodesic from $q$ to $x$. Let $v \in T_q M$ be so that $\gamma = t \in [0,1] \mapsto \exp_q(tv)$, and likewise define $v_j \in T_q M$ corresponding to $\gamma_j$. Then $v_j \to v$.

Then we can argue by contradiction: if $x$ were not conjugate to $q$ along $\gamma$, then $(\mathrm{d} \exp_q)(v): T_v( T_q M) \to T_x M$ would be invertible. But then $(\mathrm{d} \exp_q)(v_j)$ would also be invertible, once $j$ is large enough that $v_j$ lies close to $v$, and this would contradict our assumption about the $x_j$.