This can be proved with elementary linear algebra. Let us write $A_t$ for $A(t)$. Let $r$ be the constant rank of all $A_t$, where $0<r<n$. (In your case $r=n-2$.)
For any pair of matrices $M$ and $K$ we call $K$ a kernel matrix of $M$ if its columns form a basis of $\ker(M)$. In the present context, for any interval $U\subseteq\mathbb R$, we call $B:U\to M_{n,n-r}(\mathbb R)$ a kernel matrix function (of $A$) if $B_t$ is a kernel matrix of $A_t$ for every $t\in U$.
To construct a continuous kernel matrix function on $\mathbb R$, it suffices to show that given any $\tau\in\mathbb R$ and any kernel matrix $K$ of $A_\tau$, there is a continuous kernel matrix function $B$ on $[\tau-1,\tau]$ or $[\tau,\tau+1]$ such that $B_\tau=K$. Once this is done, we may start with a kernel matrix $K$ of $A_0$ and construct iteratively a continuous kernel matrix function on each of the intervals $[0,1],[1,2],\ldots$ and $[-1,0],[-2,-1],\ldots$, so that any two functions constructed on two adjacent intervals agree at the common endpoint. Then a continuous kernel matrix function on the whole real line is obtained by gluing them.
So, without loss of generality, assume that $\tau=0$ and we only need to construct a continuous kernel matrix function $B$ on $[0,1]$ with $B_0$ equals some prespecified kernel matrix $K$ of $A_0$.
Since $A_t$ has a constant rank $r$ on $[0,1]$, there exists, at any time $t$, an index set $\mathcal I(t)\subseteq\{1,2,\ldots,n\}$ of cardinality $r$ such that the columns of the submatrix $A_t(:,\mathcal I(t))$ form a basis of the column space of $A_t$. As $A$ is a continuous function, there is also an open interval $U(t)\ni t$ such that the submatrix $A_s(:,\mathcal I(t))$ has full column rank for all $s\in U(t)$.
The collection $\{U(t):t\in[0,1]\}$ is an open cover of $[0,1]$. Hence we may retrieve from it a finite subcover $\{(a_i,b_i) : 0\le i\le k\}$. By a reindexing we may assume that $a_0\le a_1\le\ldots\le a_k$. By a judicial removal of intervals, we may also assume that no interval in this subcover is a subinterval of another. It follows that $a_{i-1}<a_i<b_{i-1}<b_i$ for each $i$.
We shall construct our continuous kernel matrix function $B$ on $[0,1]$ iteratively:
- (Initialisation.) Construct $B$ on $[0,b_0)$ from a kernel matrix $K$ of $A_0$.
- (Iteration.) For $i=1,2,\ldots,k$, extend the previously constructed $B$ on $[0,b_{i-1})$ to one on $[0,b_i)$.
- (Finalisation.) Trim the domain of $B$ from $[0,b_k)$ to $[0,1]$.
In the first step, we need the following lemma, whose proof is easy and is omitted.
Lemma. Let $0<r<n$ and $M=\pmatrix{X&Y}\in M_n(\mathbb R)$ be a rank-$r$ matrix, where $X$ has $r$ linearly independent columns. A matrix $K$ is a kernel matrix of $M$ if and only if
$$
K=\pmatrix{-X^+YW\\ W}\text{ for some } W\in GL_{n-r}(\mathbb R).
$$
Similarly, if $\mathcal I$ is an index set of cardinality $r$ such that the columns of $M(:,\mathcal I)$ form a basis of the column space of $M$, and $\mathcal J$ is its complement in $\{1,2,\ldots,n\}$, then $K$ is a kernel matrix of $M$ if and only if
$$
\pmatrix{K(\mathcal I,:)\\ K(\mathcal J,:)}
=\pmatrix{-M(:,\mathcal I)^+M(:,\mathcal J)W\\ W}\text{ for some } W\in GL_{n-r}(\mathbb R).
$$
Recall that for each $i$, there is an index set $\mathcal I(t_i)$ such that the rank of $A_s(:,\mathcal I(t_i))$ is $r$ for all $s\in(a_i,b_i)$. Let us abuse the notation and write $\mathcal I_i$ for $\mathcal I(t_i)$. Let $\mathcal J_i$ be the complement of $\mathcal I_i$ in $\{1,2,\ldots,n\}$. Now suppose that a kernel matrix $K$ of $A_0$ is given. By our lemma, we must have
$$
\pmatrix{K(\mathcal I_0,:)\\ K(\mathcal J_0,:)}
=\pmatrix{-A_0(:,\mathcal I_0)^+ A_0(:, \mathcal J_0) W_0\\ W_0}
$$
for some invertible matrix $W_0$. Since the rank of $A_s(:,\mathcal I_0)$ is also $r$ for all $s\in (a_0,b_0)$. By the lemma above, the function $B$ defined on $(a_0,b_0)$ by
$$
\pmatrix{\color{red}{B_s}(\mathcal I_0,:)\\ \color{red}{B_s}(\mathcal J_0,:)}
=\pmatrix{-A_{\color{red}{s}}(:,\mathcal I_0)^+ A_{\color{red}{s}}(:, \mathcal J_0) W_0\\ W_0}
$$
is a kernel matrix function, and obviously $B_0=K$. Moreover, as the Moore-Penrose pseudo-inverse of a matrix $X$ of full column rank is $X^+=(X^TX)^{-1}X^T$, we see that $B$ is continuous on $(a_0,b_0)$. Now the initialisation step is completed by trimming the domain of $B$ from $(a_0,b_0)$ to $[0,b_0)$.
Next, in each iteration, suppose $B$ has been constructed on $[0,b_{i-1})$. Since $a_{i-1}<a_i<b_{i-1}<b_i$, there exists some positive $\tau\in(a_{i-1},b_{i-1})\cap(a_i,b_i)$. By a similar argument to the initialisation step, there is a continuous kernel matrix function $C$ on $[\tau,b_{i+1})$ such that $C_\tau=B_\tau$. A new continuous kernel matrix function $B$ on $[0,b_i)$ is then obtained by gluing the restriction of the old $B$ on $[0,\tau]$ and $C$ together. After $k$ iterations and the finalisation step, we are done.
