GP 1.2.2 $T_x(U) = T_x(X) \text{ for } x \in U.$ This is exercise 1.2.2 on Guillemin and Pollack's Differential Topology

If $U$ is an open subset of the manifold $X$, check that
  $$T_x(U) = T_x(X) \text{ for } x \in U.$$

I am fairly confused with this problem, because I found the implicit definition of tangent space at Guillemin and Pollack's Differential Topology comes not very handy for this problem. In the book, tangent space is defined by $d\phi_0$, which is not directly defined:
By definition, the tangent space of $X$ at $x$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow X$ at 0. $\phi: V \rightarrow X$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.
So here's my attempt:
$T_x(X)$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow X$ at 0. $\phi: V \rightarrow X$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.
Because $U$ is an open subset of manifold $X$, so the best linear approximation at $x$ is the same as that of $X$ around $x$. Hence, $T_x(U)$ is the image of the map $d\phi_0: \mathbb{R}^k \rightarrow \mathbb{R}^N$, where $x + d\phi_0(v)$ is the best linear approximation to $\phi: V \rightarrow U$ at 0. $\phi: V \rightarrow U$ is a local parametrization around $x$, $X$ sits in $\mathbb{R}^N$ and $V$ is an open set in $\mathbb{R}^k$, and $\phi(0) = x$.
Therefore, $T_x(U) = T_x(X) \text{ for } x \in U.$
 A: Let $ϕ:W∈R^k→X$ be a parametrization of $X$ around $x$ so that $ϕ(W)∈X$ is an open subset of the manifold. Since $U$ is open, so is $ϕ(W)∩U$, and also (since $ϕ$ is a homeomorphism) $V:=ϕ^{−1}(ϕ(W)∩U)$. Thus, $ϕ∣_V$ (the restriction of $ϕ$ to $V$) is a parametrization of $U$ around $x$; from this it follows directly that the tangent spaces are the same.
A: Let $X \subset Y$ be a submanifold and $j : X → Y$ be the inclusion map. Then
$∀x \in X, dj_x : T_x(X) → T_x(Y )$ is injective—in fact it is an inclusion. If $U$ is a open subset of a manifold $X$, $T_x(U) = T_x(X)$ for $x \in U$.
Lemma: If $X$ is a manifold, $x \in X$, and $\phi : U → X$ is a local parametrization with $\phi(0) = x$, then $(d\phi_0)^{−1} = d(\phi^{−1})_x$. Now, let $x \in X$, $\phi : U → X, \psi : V → Y$ be local parameterizations $(U \subset \mathbb R^k, V \subset \mathbb R^l$ open, $\phi(0) = x = \psi(0))$. Note that $\psi^{−1} \circ j \circ \phi = \psi^{−1} \circ \phi$, and so we have $dj_x = d\psi_0 \circ d(\psi^{−1} \circ \phi)_0 \circ (d\phi_0)^{−1} = d\psi_0 \circ d(\psi^{−1})_y \circ d\phi_0 \circ (d\phi_0)^{−1} = Id$
$\blacksquare$
