= Solution
Let $d=\dim X$ and choose a local parametrization
$$
\phi:U\subseteq\mathbb R^d\longrightarrow X,
\qquad \phi(u_0)=p,
$$
whose derivative has rank $d$. Define
$$
T_pX=\operatorname{im}D\phi_{u_0}\subseteq\mathbb R^n.
$$
This is a vector subspace because it is the image of a linear map, and rank $D\phi_{u_0}=d$ gives $\dim T_pX=d$.
If $\psi$ is another local parametrization through $p$, the transition map $h=\psi^{-1}\circ\phi$ is a local diffeomorphism. The chain rule gives
$$
D\phi=D\psi\,Dh,
$$
and $Dh$ is invertible, so $\operatorname{im}D\phi=\operatorname{im}D\psi$. Thus the definition is independent of the parametrization. Equivalently, $T_pX$ is the set of velocity vectors of smooth curves in $X$ through $p$. This is the <tangent space from a local parametrization>.
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