[proofplan]
Strict stationarity first identifies every one-dimensional marginal distribution with that of $X_0$, so all second moments and means are the same. The finite second moment also gives finite first moments and finite cross moments, so every covariance appearing in the weak stationarity condition is well-defined. Finally, strict stationarity applied to the two-dimensional vector $(X_{r+h},X_r)$ shows that its joint distribution is the same as that of $(X_h,X_0)$, and covariance is determined by this joint distribution once second moments are finite.
[/proofplan]
[step:Transfer the finite second moment and the mean to every time]
Let $\mu\in\mathbb R$ denote the mean of $X_0$:
\begin{align*}
\mu := \mathbb E[X_0].
\end{align*}
This expectation is finite because
\begin{align*}
|X_0| \leq 1 + X_0^2
\end{align*}
and $\mathbb E[X_0^2]<\infty$.
Fix $t\in\mathbb Z$. Applying strict stationarity with $n=1$, $t_1=0$, and shift $k=t$, the random variables $X_t$ and $X_0$ have the same distribution. Therefore the non-negative random variables $X_t^2$ and $X_0^2$ have the same distribution, and hence
\begin{align*}
\mathbb E[X_t^2]=\mathbb E[X_0^2]<\infty.
\end{align*}
Similarly, since $X_t$ and $X_0$ have the same distribution and $X_0$ is integrable,
\begin{align*}
\mathbb E[X_t]=\mathbb E[X_0]=\mu.
\end{align*}
Thus every $X_t$ has finite second moment and the mean is independent of $t$.
[guided]
The first weak stationarity requirement is that the first two moments exist and that the mean does not depend on time. Strict stationarity gives exactly the distributional invariance needed for this.
Define $\mu\in\mathbb R$ by
\begin{align*}
\mu := \mathbb E[X_0].
\end{align*}
This is finite because $\mathbb E[X_0^2]<\infty$ and the pointwise inequality
\begin{align*}
|X_0| \leq 1 + X_0^2
\end{align*}
implies $\mathbb E[|X_0|]<\infty$.
Now fix an arbitrary time $t\in\mathbb Z$. Strict stationarity says that shifting any finite collection of time indices does not change the joint distribution. For the one-dimensional collection consisting only of time $0$, use $n=1$, $t_1=0$, and shift $k=t$. This gives equality in distribution: $X_t$ and $X_0$ have the same distribution. Because the map $x\mapsto x^2$ is Borel measurable, the random variables $X_t^2$ and $X_0^2$ also have the same distribution.
Since $X_0^2$ is integrable by hypothesis, the expectations agree:
\begin{align*}
\mathbb E[X_t^2]=\mathbb E[X_0^2]<\infty.
\end{align*}
The same equality in distribution, now applied to the integrable function $x\mapsto x$, gives
\begin{align*}
\mathbb E[X_t]=\mathbb E[X_0]=\mu.
\end{align*}
Because $t\in\mathbb Z$ was arbitrary, every time coordinate has the same finite second moment and the same mean.
[/guided]
[/step]
[step:Verify that all cross moments and covariances are finite]
Let $r,h\in\mathbb Z$. Since $X_{r+h}$ and $X_r$ have finite second moments, the pointwise inequality
\begin{align*}
|X_{r+h}X_r| \leq \frac{1}{2}X_{r+h}^2+\frac{1}{2}X_r^2
\end{align*}
gives
\begin{align*}
\mathbb E[|X_{r+h}X_r|] \leq \frac{1}{2}\mathbb E[X_{r+h}^2]+\frac{1}{2}\mathbb E[X_r^2]<\infty.
\end{align*}
Hence the covariance
\begin{align*}
\operatorname{Cov}(X_{r+h},X_r):=\mathbb E[(X_{r+h}-\mu)(X_r-\mu)]
\end{align*}
is finite for every $r,h\in\mathbb Z$.
[/step]
[step:Use strict stationarity on pairs to make covariance depend only on the lag]
Define the autocovariance candidate $\gamma:\mathbb Z\to \mathbb R$ by
\begin{align*}
\gamma(h):=\operatorname{Cov}(X_h,X_0).
\end{align*}
Fix $r,h\in\mathbb Z$. Apply strict stationarity with $n=2$, time indices $h$ and $0$, and shift $k=r$. Then the random vectors $(X_{r+h},X_r)$ and $(X_h,X_0)$ have the same distribution. The map $m:\mathbb R^2\to\mathbb R$ given by $m(a,b)=(a-\mu)(b-\mu)$ is Borel measurable. Its absolute value is integrable under both joint laws, because both coordinates have finite second moments and
\begin{align*}
|(a-\mu)(b-\mu)| \leq \frac{1}{2}(a-\mu)^2+\frac{1}{2}(b-\mu)^2
\end{align*}
for all $a,b\in\mathbb R$. Therefore equality of the joint distributions implies equality of the expectations of $m$:
\begin{align*}
\operatorname{Cov}(X_{r+h},X_r)=\operatorname{Cov}(X_h,X_0)=\gamma(h).
\end{align*}
Thus the covariance depends only on the lag $h$.
[/step]
[step:Conclude weak stationarity]
We have shown that $\mathbb E[X_t^2]<\infty$ for every $t\in\mathbb Z$, that $\mathbb E[X_t]=\mu$ is independent of $t$, and that the function $\gamma:\mathbb Z\to\mathbb R$ satisfies
\begin{align*}
\operatorname{Cov}(X_{r+h},X_r)=\gamma(h)
\end{align*}
for all $r,h\in\mathbb Z$. These are precisely the conditions for weak stationarity. Hence $(X_t)_{t\in\mathbb Z}$ is weakly stationary.
[/step]
