Let $p: Y \to X$ be a covering with fix base point.
I have already shown that the induced homomorphism $p_{*}=\pi_1(Y,y_0) \to \pi_1(X,x_0)$ is injective.
However, since we are not calling it an isomorphism, I am going guess that such a covering map is not surjective.
Unfortunately, I cannot convince myself of this.
From the path lifting lemma, it seems that $\forall [\alpha] \in \pi_1(X,x_0)$, there exists $[\tilde{\alpha}] \in \pi(Y,y_0)$, namely the set of all loops in $Y$, $\tilde{\alpha}$, such that $p \circ \tilde{\alpha}=\alpha$ for all $\alpha \in [\alpha]$.
Then $p_{*}[\tilde{\alpha}]=[p \circ \tilde{\alpha}]=[\alpha]$.
What am I missing/not understanding about the path lifting lemma and it's relation to the fundamental group?
The lifting of a loop in $(X,x_0)$ need not be a loop. Consider $(S^1,(1,0))$ with the covering $p:\Bbb R\to S^1,\ t\mapsto (\cos2\pi t,\sin2\pi t)$. The lift of $p|_I$ at $0$ is the inclusion $I\hookrightarrow \Bbb R$.
One can show that if $[f]\in p_*(\pi_1(Y,y_0))\subseteq \pi_1(X,x_0)$, then its lift $\tilde f$ at $y_0$ must be a loop in $\pi_1(Y,y_0)$. So the generator $p|_I$ of $\pi_1(S^1,s_0)$ is not in the image of $\pi_1(\Bbb R,0)$