Your intuition is totally correct.
$v = \mathbf{1}$ is an eigenvector to eigenvalue $k$, since $A \mathbf{1} = k \mathbf{1}$.
Now, let $v_i$ denote the characteristic function of the $i$'th connected component, i.e., $v_i(x)$ is $1$ for all vertices $x$ within the $i$'th component and $0$ else, thus, it is $\mathbf{1} = \sum_{i=1}^t v_i$.
For any $i = 1, \ldots, t$ you get that $A v_i = k v_i$.
Thus, as you argued, if $X$ is not connected, then you can use the block diagonalized form to show that the $v_i$'s are $t>1$ distinct (even orthogonal) eigenvectors to the eigenvalue $k$.
Clearly, $\mathbf{1}$ is still an eigenvector, but it is composed by multiple other eigenvectors that form the $t$-dimensional eigenspace to the eigenvalue $k$.
Further, you do not even need the block diagonalized form, although it clearifies what's going on. In general, right-multiplication to the adjacency matrix with a vector $w \in \{0,1\}^n$ gives a vector that collects amount $1$ from all $1$-valued neighbors. So, the positions of the zero-entries in $A w$ match the positions of the zero-entries in $w$ if and only if $w$ is a characteristic vector of some connected component.
Best Answer
Suppose that $G$ is $d$-regular. Then every vertex of $G$ is adjacent to $d$ other vertices, so each row of its adjacency matrix $A$ will have $d$ $1$’s and $n-d$ $0$’s. Let
$$A=\pmatrix{a_{11}&a_{12}&\dots&a_{1n}\\ a_{21}&a_{22}&\dots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{n1}&a_{n2}&\dots&a_{nn}}\;,$$
and $\vec v$ be the $n\times 1$ vector whose entries are all $1$:
$$\vec v=\pmatrix{1\\1\\\vdots\\1}\;.$$
The product $\vec u=A\vec v$ is an $n\times 1$ vector whose $i$-th entry is $$u_i=\sum_{j=1}^na_{ij}\cdot1=a_{i1}+a_{i2}+\ldots+a_{in}\;.$$ Since $d$ of the numbers $a_{i1},\dots,a_{in}$ are $1$ and the rest are $0$, $u_i=d$ for every $i$. Thus, $$\vec u=\pmatrix{d\\d\\\vdots\\d}=d\vec v\;.$$ That is, $A\vec v=d\vec v$, so by definition $\vec v$ is an eigenvector of $A$ for the eigenvalue $d$.
To show the other direction, you can try to reverse this argument; that works. Alternatively, you can assume that $G$ is not $d$-regular and show that $\vec v$ is not an eigenvector for an eigenvalue $d$; that also works and may be even easier.