I tried my best to be very explicit with the details here. Please let me know if any place needs more exposition.
Simple case: we're lucky and $A$ and $X$ commute.
Assume that $X$ commutes with $A$. Then, they are simultaneously diagonalizable as $X=UL_XU^T$ and $A=UL_AU^T$ for an orthogonal matrix $U$ and diagonal matrices $L_X$ and $L_A$ of the eigenvalues. The minimization problem in this case is:
$$\min\limits_{X\in\Omega}\frac{1}{2}\|UL_XU^T-UL_AU^T\|_F^2$$
but, since $U$ is an orthogonal matrix and the Frobenius norm is invariant under orthogonal transforms, we can also write
$$\|UL_XU^T-UL_AU^T\|_F^2 = \|L_XU^T-L_AU^T\|_F^2 = \|L_X-L_A\|_F^2$$
so that our problem is equivalent to
$$\min\limits_{X\in\Omega}\frac{1}{2}\|L_X-L_A\|_F^2.$$
Taking into account that $L_X$ and $L_A$ are $n\times n$ diagonal matrices (and thus identified with vectors in $\mathbb{R}^n$), the constraint $X\in\Omega$ is equivalent to $L_X\in\mathbb{R}^n_+$ (this is the psd part of the constraint) with at most $k$ nonzero components (this is the rank-$k$ part of the constraint). But, this new optimization problem is actually easily solved: we just project the vector $L_A$ onto $\mathbb{R}^n_+$ and keep only the fist $k$ nonzero components (we assume the components are sorted according to magnitude, otherwise instead of keeping the first $k$ components we would keep the $k$ most positive (largest) components).
To summarize, letting $L_A = (\sigma_1, \sigma_2, \ldots, \sigma_n)$, we take $L_X = (R(\sigma_1),R(\sigma_2),\ldots,R(\sigma_k),0,\ldots,0)$ where $R(x)=\max(0,x)$ is the ReLU function and our rank-$k$ psd solution is then $UL_XU^T$.
General case: what if we're not lucky?
The question remains: if $X$ does not commute with $A$ what can we do? It turns out that considering the case where they commute is sufficient, due to a trick/inequality related to the Frobenius norm.
von-Neumann's trace inequality says, for $A$ real and symmetric and $X$ real psd with spectral decompositions $A=VL_AV^T$ and $X=UL_XU^T$ for $U$ and $V$ orthogonal matrices and $L_A$ and $L_X$ diagonal matrices as before,
$$\|X-A\|_F^2 \geq \|L_A-L_X\|_F^2$$
Now, assume that there is a minimizer $Y\in\Omega$ of our original problem which does not commute with $A$. The spectral decomposiiton of $Y$ will be written $WL_YW^T$ for some orthogonal matrix $W$. Then we have
$$\|Y-A\|_F^2 \geq \|L_Y-L_A\|_F^2 = \|VL_YV^T - A\|_F^2$$
where the first inequality is due to the von-Neumann trace inequality and the second equality is the invariance of the Frobenius norm with respect to orthogonal transformations. The matrix $Z=VL_YV^T$ is in $\Omega$ since $Y\in\Omega$.
Thus, given some minimizer $Y$ which is not commuting with $A$, we can always construct the matrix $Z\in\Omega$ which does commute with $A$ and which is at least as good in terms of the objective function:
$$\|Y-A\|_F^2\geq \|Z-A\|_F^2$$