Let $G$ have $\ell$ components $C_1$, $\ldots\,$, $C_{\ell}$. For $i = 1$, $\ldots\,$, $\ell$, define the vector $x^{(i)}$ by $x^{(i)}_j = \chi_{v_j \in C_i}$.
Since the different $x^{(i)}$s are orthogonal, in order to show that the multiplicity of $k$ is $\ell$, it is enough to show that each $x^{(i)}$ is an eigenvector with eigenvalue $k$ and that if $x$ is an eigenvector with eigenvalue $k$ then there exist $\lambda_1$, $\ldots\,$, $\lambda_{\ell}$ (with some $\lambda_i \neq 0$) such that $x = \lambda_1 x^{(1)} + \cdots + \lambda_{\ell} x^{(\ell)}$.
The first assertion is easy to see. For the second, let $x$ be an eigenvector with eigenvalue $k$. Without loss of generality, assume that $\lvert x_1 \rvert$ is maximal, that $x_1 > 0$, and that $v_1 \in C_1$.
Because $x$ is an eigenvector with eigenvalue $k$, we have $k x_1 = \sum_{(v_1,v_j)\in E} x_j.$ Furthermore, by our hypotheses that $G$ is $k$-regular and that $\lvert x_1 \rvert$ is maximal, we have $\sum_{(v_1,v_j)\in E} x_j \leq k \max_{(v_1,v_j)\in E} \lvert x_j \rvert \leq k x_1.$ Note that because $G$ is $k$-regular, the sum on the left-hand side has exactly $k$ non-zero terms. Hence, all of them are positive and all equal $x_1$.
We can apply the same reasoning to each $x_j$ for which $(v_1,v_j)\in E$, and so on to all of the $x_i$ for which $v_i \in C_1$.
It may be that there exists some $j \in [n]$ such that $v_j \notin C_1$ but $x_j \neq 0$. However, we can apply the argument used above to show that for each $i$, if $v_j \in C_i$, then $x_j = \max_{v_j \in C_i} \lvert x_j \rvert$. Thus, we have shown that $x$ is a linear combination of the $x^{(i)}$s, as claimed.