Demonstrating that probability for every possible result is uniform at the end of an algorithm
The proof is by induction. The base case $t = k$ is clear. Suppose that the claim is true at some time $t$. We will prove it for time $t+1$.

Let the first $t+1$ elements be $x_1,\ldots,x_{t+1}$. By the induction hypothesis, at time $t$ each of the $\binom{t}{k}$ possible $k$-subsets of $x_1,\ldots,x_t$ is found in the array with equal probability. The probability that at step $t+1$ the array remains the same is $1-k/(t+1)$, hence each of the $k$-subsets of $x_1,\ldots,x_t$ appears at time $t+1$ with probability
$$
\frac{t+1-k}{t+1} \frac{1}{\binom{t}{k}} = \frac{1}{\binom{t+1}{k}}.
$$
Now consider some $k$-subset $S$ of $x_1,\ldots,x_{t+1}$ that contains $x_{t+1}$. For this set to be appear at time $t+1$, the following two events need to happen: at time $t$, the array consisted of $S \setminus \{x_{t+1}\}$ together with one of the $t-(k-1)$ remaining elements; and at time $t+1$, this element was replaced by $x_{t+1}$. In total, the probability is
$$
\frac{k}{t+1} \cdot \frac{1}{k} \cdot \frac{t-k+1}{\binom{t}{k}} = \frac{1}{\binom{t+1}{k}}.
$$