Is this a legit way to prove that a fundamental group of a connected graph $\Gamma$ is a free group? Without using quotient and homotopy extension property from Hatcher's "Algebraic Topology":
Take a maximal tree $T \subset \Gamma$, built on a vertex $v_0$. Take a set of edges $\{ e_\alpha | \alpha \in A\}$ s.t. $e_ \alpha \in\Gamma \setminus T$.
First, suppose $A=\{1\}$. Then there is a unique reduced cycle in $\Gamma$ containing $e_\alpha$. Call it $L$. As a loop subgraph, $L \cong S^1$ (homotopy equivalent).
Now we prove that $\Gamma \cong L$: all components of $\overline{\Gamma \setminus L}$ are trees, and if any component intersects $L$ in more than one point, then it either forms a second reduced cycle in $\Gamma$ containing $e_\alpha$, or we get a reduced cycle in $T$. Both are a contradiction, so each component of $\overline{\Gamma \setminus L}$ has a unique intersection point with $L$, and since $T$ is a tree, we can deformation retract this component to its unique intersection point. So $\Gamma$ deformation retracts to $L$, and $\Gamma \cong L$.
Now for general $A$, let $m_ \alpha$ be some midpoint of $e_\alpha$. Let $\Gamma_\alpha= (\Gamma \setminus \cup_{\beta \in A} m_\beta) \cup m_\alpha$. From case 1 we have that $\Gamma_\alpha$ has a fundamental group $\mathbb Z$. And $\Gamma _ \alpha$ satisfy the conditions of Seifert-Van Kampen theorem: open, path-connected, intersection non-empty, double and triple intersections simply connected. Then $\pi _1 (\Gamma) \cong *_{\alpha \in A} \mathbb Z \cong F(e_\alpha, \alpha \in A)$ – free group on $e_\alpha$.
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Your answer seems to me to be correct. But I think it is helpful also to see this question in the broader context of free groupoids, which were introduced by Philip Higgins, see his 1971 text reprinted as Categories and Groupoids.
The relation with topology is also covered in Topology and Groupoids, as it was in the 1968 edition of this book, "Elements of Modern Topology". A graph can be regarded as a quotient of a disjoint union of unit intervals $[0,1]$ by identifying various vertices. This gives a $1$-dimensional CW-complex. Now the fundamental groupoid $\pi_1([0,1], \{0,1\})$ on the two base points $0,1$ is a groupoid which we sometimes write $\mathcal I$; a free groupoid is obtained from a disjoint union of copies of $\mathcal I$ by identifying some of the vertices. This is a special case of a groupoid construction in which given a groupoid $G$ with object set $X$ and a function $f:X \to Y$ to a set $Y$ one obtains a groupoid $U_f(G)$, which I also like to write $f_*(G)$, with object set $Y$ and a ``universal morphism'' $G\to U_f(G)$.
Any connected groupoid $H$ can be written as $H(a) \ast T$ where $H(a)$ is the vertex group of $H$ at the object $a$ and $T$ is a "tree groupoid'', i.e. each $T(x,y)$ has exactly one element. If $H$ is a free groupoid, then $H(a)$ is a free group.
See this stackexchange answer for a result on free groupoids and an application of the groupoid Seifert-van Kampen Theorem.