I followed these steps, but the answer still says that this system is time-invariant
let: $x_2[n] = x[n-k]$
$$\begin{align} y_2[n] &= x_2[-n] \\ &= x[-(n-k)] \\ &= x[k-n] \\ \end{align}$$
and
$$\begin{align} y[n-k] &= x[(-(n-k)] \\ &= x[k-n] \end{align}$$
where am I wrong in the analysis?
A time-invariant system is one that, when you shift the input signal, the output is shifted by the same amount.
A system that reverses the signal cannot be time-invariant because when you shift the input, the output is shifted the other way. $k$ and $-k$ are not the same amount.
$$ y[n-k] = x[k-n] = x[-n \mathbf{+} k] $$
I would suggest to first try with simple examples, to get more insight. And a counter example might suffice to disprove the claim.
Here, take the discrete delta $\delta_k(n)$, zero everywhere, except at $n=k$, where $\delta(k)=1$. This discrete input is the paragon for time-invariant linear systems.
So: with your system $\mathcal{S}$, you get $$\mathcal{S} \delta_0(n) = \delta_0(-n)= \delta_0(n)$$ a mirror-symmetry invariant. But $$\mathcal{S} \delta_1(n) = \delta_{-1}(n)$$
So, time-invariant here is doomed.
You have asked what is wrong in your analysis. First let me explain the intuition: You have a sequence X(n). At n=1 the sequence takes the value X(1). At n=2 the sequence takes X(2) and so on. You now define Y(n) as X(-n). This means at n=1, Y(n) takes the value of X(-1), at n=2, Y(n) takes the value X(-2) and so on. Now, you delay the input sequence by k. This means at n=1 the value of the sequence is X(1-k) at n=2 the value is X(2-k) and so on. Now for the same system when this is the input sequence, at n=1, Y(1) = the value of the input sequence at n=-1 which is Y(-1-k). Now can you see the big picture?
There are many ways to understand this: When you take X(n-k) and then flip it as X(-(n-k)) then you are reflecting the input sequence about X=k. But the system reflects the input sequence around X=0.
Now to answer, your question about where your analysis is wrong, let Z(n) = X( n-k). Then when this is passed to the system the output Y(n) is Z(-n) which is X(-n-k) and not X(-(n-k)) which is X(-n+k).
It is difficult to wrap one's head around this. Atleast it was so for me. But hope it helps.
The point is even if the result is the same in both when the input is x[n-k] and when the output is y[n-k], the output x[-n+k] is shifted to the left, whilst the input is shifted to the right. So if the system was to be invariant, y[n-k] should have yielded x[-n-k], but it doesn't, it yields x[-n+k].
