If a covering map has a section, is it a $1$-fold cover? If $q: E\rightarrow X$ is a covering map that has a section (i.e. $f: X\rightarrow E, q\circ  f=Id_X$) does that imply that $E$ is a $1$-fold cover?
It follows from your assumptions that $q$ is a 1-sheeted and is a homeomorphism.  I'm going to call the map $\pi$ instead of $q$ for the rest of this post.

Assume we have a covering $\pi:X\rightarrow Y$ and $f:Y\rightarrow X$ with $\pi\circ  f = Id_X$.

I claim that $f(Y)$ is both open and closed in $X$.

To see it, for any $\hat{p}\in X$, let $p = \pi(\hat{p})$.  Choose a neighborhood $U$ around $p$ for which $\pi$ trivializes: $\pi^{-1}(U) = \coprod V_\alpha$ with $\pi|_{V_\alpha}$ a homeomoprhism. and let $V$ be the particular $V_\alpha$ containing $\hat{p}$.

Now, if $\hat{p}\in f(Y)$, then $V\subseteq f(Y)$.  This follows from considering the inclusion $i:U\rightarrow Y$.  Since both $f|_{U}$ and $\pi^{-1}|_{U}$ are lifts of this inclusion agreeing at $\hat{p}$, they must agree on all of $U$.  It follows that $V=\pi^{-1}(U) = f(U)$ as claimed.  This shows $f(Y)$ is open.

If, on the other hand $\hat{p}\notin f(Y)$, a very similar argument shows that $V\cap f(Y) = \emptyset$, showing that $f(Y)^c$ is open, that is, that $f(Y)$ is closed.

Putting this together, $f(Y)$ is open and closed.  Hence, it is a connected component of $X$.  If $X$ itself is connected, this implies $f(Y) = X$ which implies that $\pi$ is a homeomorphism with inverse $f$ so, is in particular, 1 sheeted.
A connected covering space $f:E\rightarrow X$ admits no section ( global section) unless $f$ is a homeomorphism.

Edit: looking @Andy's post I'm not so sure of what I said now.