Let's use a new concept **oriented path** in a cycle $C$ or an open ear $P_k$. <br>
1. How to construct it?

$C$: Label the vertices in $C$ in *clockwise* direction by $u,v_1,\cdots,v_n$. Then the clockwise-oriented path of $C$ ($c_1$ path) is $uv_1\cdots v_n$, and the counterclockwise-oriented path of $C$ ($c_2$ path) is $uv_n\cdots v_1$. <br>
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$P_k$: Since $P_k$ is open, it has two distinct vertices $x_1$ and $x_2$ intersecting with $\begin{cases} C \cup \bigcup_{j=1}^{k-1}P_j & k \geq 2 \\ C & k=1\end{cases}$. WLOG assume that by following $P_k$ in clockwise direction from $x_1$, we can reach $x_2$. Then label the vertices in $P_k$ in clockwise direction by $x_1,y_1,\cdots,y_\ell,x_2$. Then $c_1$ path of $P_k$ is $x_1y_1\cdots y_\ell$, and $c_2$ path of $P_k$ is $x_2y_n\cdots y_1$. <br>
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Now, we will construct two graphs $T_1$ and $T_2$: $T_1$ is a union of all $c_1$ paths of $C,P_1,\cdots,P_k$, and $T_2$ is a union of all $c_2$ paths of $C,P_1,\cdots,P_k$. And let's denote this event $E$. <br>

1. $T_1$ and $T_2$ are connected. We deleted just one edge from a cycle or an ear to construct $c_1$ and $c_2$ path. It cannot disconnect $G$. Also, by the same reason, $T_1$ and $T_2$ are acyclic.

2. $V(T_1)=V(T_2)=V(G)$.

Thus, $T_1$ and $T_2$ are spanning tree of $G$. <br>
We will use induction on $\vert V(G) \vert + \vert E(G) \vert$. <br>

Basis: The smallest $2$-connected graph is $G=K_3$ with $\vert V(G) \vert + \vert E(G) \vert=6$. By applying $E$, $T_1=uv_1v_2$ and $T_2=uv_2v_1$. Obviously, it satisfies the given statement. 

Induction hypothesis: Any graph $H$ with $\vert V(H) \vert + \vert E(H) \vert \leq m-1$ has two spanning trees obtained by $E$ such that for any $v \in U - \{u\}$, **two** $uv$-paths are independent. Note that since tree has a unique path between any pair of vertices, so we can assure that there are exactly two independent $uv$-paths, one from $T_1$ and another from $T_2$. <br>

Now, for $2$-connected $G$, there is an open ear decomposition $(C,P_1,\cdots,P_k)$ of $G$. Let's delete edges in $P_k$ from $G$, and then delete isolated vertices. Let $G'$ be the resulting graph. By hypothesis, $G'$ has two spanning trees $T_1'$ and $T_2'$ made by $E$, containing two independent $uv'$-paths for any $v' \in V(G')-\{u\}$.

Since $P_k$ is an open ear, it has at least two vertices $x_1$ and $x_2$ intersecting with $G'$. As we did above, WLOG assuem that we can reach $x_2$ from $x_1$ by following $P_k$ in clockwise direction. If $P_k$ is an edge, $T_1'$ and $T_2'$ are the desired trees. So assume $V(P_k) \geq 3$. Label the vertices in $P_k$ by $y_1,\cdots,y_n$ in clockwise direction. Now, by hypothesis, there are $ux_1$-path in $T_1'$ and $ux_2$-path in $T_2'$. Let $T_1=T_1' \cup x_1y_1\cdots y_\ell$ and $T_2=T_2' \cup x_2y_\ell\cdots y_1$. Then for any vertex $y_t$ in $P_k$, there are two independent $uy_t$-paths one from $T_1$ and another from $T_2$. Let's denote them $p_1$ and $p_2$ respectively.

1. Obviouslty, $p_1$ and $p_2$ does not intersect in $P_k$. 
2. If they intersect at $z \in V(G')$, then among $p_1 \cup p_2$, two cycles appear. Let's denote $C_1$ a cycle containing $u$, and $C_2$ a cycle containing $y_t$. Since $C_2$ should be an open ear, there is another ear (or a union of ears) $P'$ having a path $q=zz_1\cdots z_s$ in common with $C_2$. WLOG assume that $q$ is in $T_1$. Then $q$ has a clockwise orientation $zz_1\cdots z_s$. Here, observe that we must attach $P'$ to $C_1$ before attaching $C_2$ to $C_1$. But then we have to give $q$ an orientation $z_s \cdots z_1z$. It's contradiction.

As a conclusion, $p_1$ and $p_2$ are independent. Now we obtained the desired $T_1$ and $T_2$ in $G$.