= Solution
Let $B=A^T$ and choose any column <probability vector> $q$, for example the first coordinate <unit vector>. Because $A$ has nonnegative entries and each row sums to one, $B$ maps the <probability simplex>
$$
\Delta=\{v\in\mathbb R^n:v\geq0,\ \mathbf1^Tv=1\}
$$
into itself: $Bv\geq0$ and $\mathbf1^TBv=(A\mathbf1)^Tv=\mathbf1^Tv=1$.
The iterates $B^jq$ need not converge, so take their arithmetic averages,
$$
x_N=\frac1N\sum_{j=0}^{N-1}B^jq\in\Delta.
$$
A telescoping <sum> gives
$$
Bx_N-x_N=\frac{B^Nq-q}{N}.
$$
Both <vectors> on the right before subtraction have <L1 norm> one, so $\|Bx_N-x_N\|_1\leq2/N\to0$.
The <probability simplex> is closed and bounded in finite-dimensional <Euclidean space>, hence compact by the <Heine-Borel theorem>. It follows that a <subsequence> $x_{N_j}$ converges to some $x\in\Delta$. Passing to the <limit> in the displayed identity, using <continuity> of the <linear map> $B$, gives \b[$A^Tx=Bx=x$, with $x\geq0$ and $\sum_i x_i=1$].
This proves <stationary distribution existence for a finite Markov chain> without any <Markov chain irreducibility> or <aperiodicity> assumption. Those stronger hypotheses concern uniqueness and convergence, not the existence required here.
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