In Vector Bundles and K-Theory (version 2.2, p.8), Hatcher explains:
Lemma 1.1. A continuous map $h:E_1\to E_2$ between vector bundles over the same base space $B$ is an isomorphism if it takes each fiber $p_1^{-1}(b)$ to the corresponding fiber $p_2^{-1}(b)$ by a linear isomorphism.
Proof: The hypothesis implies that $h$ is one-to-one and onto. What must be checked is that $h^{-1}$ is continuous. This is a local question, so we may restrict to an open set $U\subset B$ over which $E_1$ and $E_2$ are trivial. Composing with local trivializations reduces to the case that $h$ is a continuous map $U\times\mathbb{R}^n\to U\times\mathbb{R}^n$ of the form $h(x,v)=(x,g_x(v))$. Here $g_x$ is an element of the group $GL(n,\mathbb{R})$ of invertible linear transformations of $\mathbb{R}^n$, and $g_x$ depends continuously on $x$. This means that if $g_x$ is regarded as an $n\times n$ matrix, its $n^2$ entries depends continuously on $x$. The inverse matrix $g_x^{-1}$ also depends continuously on $x$ since its entries can be expressed algebraically in terms of the entries of $g_x$, namely $g_x^{-1}$ is $1/(\det g_x)$ times the classical adjoint matrix of $g_x$. Therefore $h^{-1}(x,v)=(x,g_x^{-1}(v))$ is continuous. $\square$
The part that confuses me is in bold. Why does $g_x$ depends continuously on $x$?
Let $(e_1,…,e_n)$ be the canonical basis of $\Bbb R^n.$ The $k$-th column of $g_x$ is $g_x(e_k),$ and $x↦h(x,e_k)$ is continuous.