Wrong solution. See James' below. I'll just add that to show independence for say $X_{\sigma(1)},X_{\sigma(2)}$, $E(E(f(X_{\sigma(1)}) g(X_{\sigma(2)})|X_{\sigma(1)})) = E( f(X_{\sigma(1)}) E(g(X_{\sigma(2)}| X_{\sigma(1)}))$. It would suffice to show $E(g(X_{\sigma(2)})|X_\sigma(1)) = E(g(X_1))$. This can be done by conditioning on $\sigma(2) = j$ for $j > \sigma(1)$, and then sum over such $j$'s.

I think Ori's suggested approach is best. Let $\sigma(i)$ be the index of the $i$th k that is $1$. Then condition on $\sigma(1), \ldots, \sigma(N)$, $X_{\sigma(1)}, \ldots, X_{\sigma(N)}$ are iid. This you can easily check by computing $P(X_{\sigma(i} \in A_i)$ using tower property of conditioning.

So now you can just do a conditional LLN. By your assumption, for any $N$, $P(\sum_{j=1}^T k_j > N)$ with high probability, for sufficiently large $T$. So

$$P(\sum_{j=1}^T k_j X_j > c \sum_{j=1}^T k_j) < P(\sum_{i=1}^{\sigma^{-1}(T)} X_{\sigma(i)} > c \sigma^{-1}(T) | \sigma^{-1}(T) > N) + P(\sigma^{-1}(T) < N),$$

where $\sigma^{-1}(T)$ is the number of nonzero $k_j$'s for $j \le T$.

You can deal with the first component using Chebyshev by breaking it into an infinite sum conditioning on $\sigma^{-1}(T) = N+k$ for $k \in \mathbb{N}$, which are uniformly small; then apply Bayes' formula. The second piece is small as we discussed. The whole thing is then small.

Edit: the following earlier approach seems useless.

First of all assuming $X_j$'s are centered, your sequence $N_s := \sum_{t=1}^s k_t X_t$ is a Martingale because $E[ k_s X_s | \mathcal{F}_{s-1}] = 0$, where I let $\mathcal{F}_s$ be the sigma field generated by $X_1, \ldots, X_s, k_1, \ldots, k_s$. Thus
$$ var N_s = \sum_{j=1}^s E (k_j X_j)^2 = \sum_{j=1}^s E(k_j^2) E(X_j^2)$$ assuming $X_j$'s are centered, and using independence of $X_j$ with $k_j$. You should then be able to use Chebyshev as in the usual LLN to conclude.