Let $F$ be the finitely generated free group, say with $n$ generators, so that if $X$ is the wedge of $n$ circles with wedge point $x_0$, $\Pi_1(X,x_0)\cong F$. We may then construct the covering space $\tilde{X}$ corresponding to $N$, which we know to be a connected graph, where each vertex projects to the wedge point and each edge projects to one of the circles in $X$. As usual, we may take $T$ a maximal tree inside of $\tilde{X}$ and use the fact that the quotient $\tilde{X}/T$ is homotopoy equivalent to $\tilde{X}$, and so has the same fundamental group, namely $N$.
Everything we have done so far may be done in complete generality fo any $N\leq \Pi_1(X)$. We now prove that if $N$ is as given, it is not finitely generated.
The first step is to recall that the fundamental group of $\tilde{X}/T$ is always free with cardinality equal to the cardinality of edges in $\tilde{X}\setminus T$ (since the quotient collapses all of the verticies and edges inside $T$ down to a single point, so all that is left is a wedge of circles, with each circle corresponding to an edge not in $T$). So if $N$ were finitely generated, there would be only finitely many edges in $X\setminus T$. We will now show that if $N$ is also normal and non-trivial and of infinite index, then $T$ contains a loop, contradicting the fact that $T$ is a tree, so we win.
Since $N$ is normal, for any vertex $v \in\tilde{X}$, $p_*\Pi_1(\tilde{X},v)=N$, and since $N$ is non-trivial, there is some non-trivial loop $\gamma$ in $X$ based at $x_0$ corresponding to an element of $N$. This may be expressed in terms of a reduced word, say of length $k\geq 1$. (We count e.g. $a^2$ as having length $2$.) We will find a vertex $v$ such that the lift $\tilde{\gamma}$ of $\gamma$ at $v$ lies inside of $T$. But then since $[\gamma] \in p_*\Pi_1(\tilde{X},v) = N$ is non-trivial, $\tilde{\gamma}$ is a non-trivial loop in $T$. (It is a loop since $\gamma$ lies in $p_*\Pi_1(\tilde{X},v)$).
To find $v$, let $V_0$ be the set of verticies such that one of the edges of that vertex lies outside of $T$. Let $V_i$ be the set of verticies of distance at most $i$ from some element of $V_0$ (i.e. the set of verticies that can be reached from following a path starting at $V_0$ that travels no more than $i$ edges). Since $N$ is of infinite index, $\tilde{X}$ has infinitely many verticies, but since each vertex has finitely many edges (each vertex must have one arrow going in and one going out for each generator of $F$, so has at most $2n$ edges) $V_i$ is always a finite set. Thus we may take any $v \in \tilde{X} \setminus V_{k}$. Since $\gamma$ is a word of length $k$, the lift of $\gamma$ at $v$ is a path of length $k$ and so never hits a vertex in $V_0$. But this means that all of the edges travelled by $\tilde{\gamma}$ lie inside of $T$, as claimed.