Disclaimer: This is, so far, one of my most downvoted answers on the site. Needless to say, it is perfectly correct, and it answers the question as formulated at the time. I guess one should consider such erratic downvotes as an inherent part of the math.SE experience... Happy reading!
This is to explain how to show the result with no Fourier analysis, using only infinite sequences of heads and tails, so familiar to probabilists, and to draw a parallel with the perhaps better known doubling map.
First, the doubling map. As is well known, the unit interval $[0,1]$ is bimeasurably equivalent, up to null sets for the Lebesgue measure, to the infinite dimensional discrete cube $\{0,1\}^\mathbb N$ by the dyadic mapping $x\mapsto\epsilon=(\epsilon_n)_{n\geqslant1}$ defined, up to null sets for the Lebesgue measure, by
$$
x=\sum_{n=1}^\infty\frac{\epsilon_n}{2^n}.
$$
Then the doubling map $B:[0,1]\to[0,1]$ defined by $B(x)=2x\bmod{1}$, corresponds to the shift $\beta$ defined on $\{0,1\}^\mathbb N$ by
$$
(\beta\epsilon)_n=\epsilon_{n+1}
$$
for every $n$. Thus, for every nonnegative $k$, $$
(\beta^k\epsilon)_n=\epsilon_{n+k}
$$
for every $n$, $B$ leaves invariant the Lebesgue measure on $[0,1]$ and $\beta$ leaves invariant the product uniform measure on $\{0,1\}^\mathbb N$. Furthermore, if $x$ is uniform on $[0,1]$ then $(\epsilon_n)_{n\geqslant1}$ is i.i.d. Bernoulli uniform on $\{0,1\}$ hence, by Kolmogorov zero-one law, the tail sigma-algebra
$$
\bigcap_{k\geqslant0}\sigma(\beta^k)=\bigcap_{k\geqslant0}\sigma(\epsilon_n;n\geqslant k)
$$
is trivial, that is, $\beta$ is ergodic for the uniform measure on $\{0,1\}^\mathbb N$ hence, equivalently, $B$ is ergodic for the Lebesgue measure on $[0,1]$.
Now, to the tent map $T:[0,1]\to[0,1]$ defined by $T(x)=\min(2x,2-2x)$. This time, we encode every $x$ in $[0,1]$ by some sequence $\eta=(\eta_n)_{n\geqslant1}$ in $\{-1,1\}^\mathbb N$ (the replacement of $\{0,1\}$ by $\{-1,1\}$ being merely cosmetic) through the identity
$$
x=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_n}{2^n}.
$$
To get some familiarity with this somewhat non standard expansion, note that the intervals $(0,\frac12)$, $(\frac12,1)$, $(0,\frac14)$, $(\frac14,\frac12)$, $(\frac12,\frac34)$ and $(\frac34,1)$ correspond to $\eta_1=1$, to $\eta_1=-1$, to $\eta_1=\eta_2=1$, to $\eta_1=-\eta_2=1$, to $\eta_1=-\eta_2=-1$ and to $\eta_1=\eta_2=-1$ respectively.
A key fact is that, if $x$ is uniform on $[0,1]$ then $(\eta_n)_{n\geqslant1}$ is (again) i.i.d. Bernoulli uniform (but this time) on $\{-1,1\}$. Furthermore, the map $T$ corresponds to a twisted shift $\vartheta$ defined by $$(\vartheta\eta)_n=\eta_1\eta_{n+1}$$ for every $n$. The proof is direct: $x<\frac12$ corresponds to $\eta_1=1$, then $$
x=\frac14-\frac12\sum_{n=2}^\infty\frac{\eta_n}{2^n}
$$
hence
$$
Tx=2x=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_{n+1}}{2^n}=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_1\eta_{n+1}}{2^n}
$$
Likewise, $x>\frac12$ corresponds to $\eta_1=-1$, then
$$
x=\frac34-\frac12\sum_{n=2}^\infty\frac{\eta_n}{2^n}
$$
hence
$$
Tx=2-2x=\frac12+\frac12\sum_{n=1}^\infty\frac{\eta_{n+1}}{2^n}=\frac12-\frac12\sum_{n=1}^\infty\frac{\eta_1\eta_{n+1}}{2^n}
$$
Now, iterating the shift $\vartheta$ is simple, since, for every nonnegative $k$, $$(\vartheta^k\eta)_n=\eta_k\eta_{n+k}$$ for every $n$, thus, $$
\bigcap_{k\geqslant0}\sigma(\vartheta^k)\subseteq\bigcap_{k\geqslant0}\sigma(\eta_n;n\geqslant k)
$$
is trivial (and the inclusion is an identity), that is, $\vartheta$ is ergodic for the uniform measure on $\{-1,1\}^\mathbb N$ hence, equivalently, $T$ is ergodic for the Lebesgue measure on $[0,1]$.
Best Answer
Isomorphisms preserve ergodicity. So if you can construct an isomorphism of the baker map to a bernoulli shift you are done. I think this is the standard way it is done in most books.