@Anne has answered my question in a comment. I formalize her ideas as below.
We have $L^\infty(Y)=(L^1(Y))^*$, so
$$
\|\pi_\sharp f\|_{L^\infty(Y)} = \sup \{\langle \pi_\sharp f, g \rangle_{L^1(Y)} ; \|g\|_{L^1(Y)} = 1\}.
$$
We have the duality
$$
\langle \pi_\sharp f, g \rangle_{L^1(Y)} = \langle f, \pi^\sharp g \rangle_{L^1(X)}.
$$
So
$$
\|\pi_\sharp f\|_{L^\infty(Y)} = \sup \{\langle f, \pi^\sharp g \rangle_{L^1(X)} ; \|g\|_{L^1(Y)} = 1\}.
$$
Notice that
$$
\begin{align}
\langle f, \pi^\sharp g \rangle_{L^1(X)} &\le \|f\|_{C(X)} \langle \pmb 1, \pi^\sharp g \rangle_{L^1(X)} \\
&= \|f\|_{C(X)} \int_X \pi^\sharp g \mathrm d \mu \\
&= \|f\|_{C(X)} \int_X g \circ \pi \mathrm d \mu \\
&= \|f\|_{C(X)} \int_Y g \mathrm d \nu \quad (\star)\\
&\le \|f\|_{C(X)} \|g\|_{L^1(Y)}.
\end{align}
$$
Here $\pmb 1$ is the constant map $x \mapsto 1$. Also, $(\star)$ follows from change-of-variables formula. This completes the proof.
Update 1: @Anne has pointed out a mistake in above proof, i.e., the duality holds for $g \in L^2(Y)$ but not necessarily for $g \in L^1(Y)$. However, it seems we can obtain the same inequality with a tweak on the definition of $\pi^\sharp$. We define a map
$$
\pi^\sharp: L_1(Y) \to L_1(X), g \mapsto g \circ \pi.
$$
Then $\pi^\sharp$ is a linear operator. Moreover,
$$
\|\pi^\sharp\| = \sup_{g \in L_1(Y)} \frac{\|g \circ \pi\|_{L_1(X)}}{\|g\|_{L_1(Y)}} = \sup_{g \in L_1(Y)} \frac{\int_X |g| \circ \pi \mathrm d \mu}{\int_Y |g| \mathrm d \nu} = \sup_{g \in L_1(Y)} \frac{\int_Y |g| \mathrm d \nu}{\int_Y |g| \mathrm d \nu} = 1.
$$
It follows that $\pi^\sharp$ is bounded. We denote its adjoint by
$$
\pi_\sharp : L_\infty(X) \to L_\infty(Y).
$$
We have the duality
$$
\int_Y (\pi_\sharp f) g \mathrm d \nu = \int_X f (\pi^\sharp g) \mathrm d \mu \quad \forall f \in L_\infty(X), \forall g \in L_1(Y). \quad (\star)
$$
For $f \in C(X)$, we have $f \in L_\infty(X)$, and
$$
\begin{align}
\|\pi_\sharp f\|_{L_\infty(Y)} &= \sup \left \{ \int_Y (\pi_\sharp f) g \mathrm d \nu \,\middle\vert\, \|g\|_{L_1(Y)} = 1 \right\} \\
&= \sup \left \{ \int_X f(\pi^\sharp g) \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \quad \text{by } (\star)\\
&\le \|f\|_\infty \sup \left \{ \int_X \pi^\sharp g \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\
&= \|f\|_\infty \sup \left \{ \int_X g \circ \pi \mathrm d \mu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\
&= \|f\|_\infty \sup \left \{ \int_Y g \mathrm d \nu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \quad (\star\star)\\
&\le \|f\|_\infty \sup \left \{ \int_Y |g| \mathrm d \nu \,\middle\vert\, \|g\|_{L^1(Y)} = 1 \right\} \\
&= \|f\|_\infty.
\end{align}
$$
Here $(\star\star)$ follows from change-of-variables formula. This completes the proof.
Update 2: I show here why Tao's function $\pi_\sharp f\in L_2(Y)$ is equal $\nu$-a.e. to my function $\pi'_\sharp f\in L_\infty(Y)$. We have
$$
\int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y\left(\pi_{\sharp} f\right) g \mathrm d \nu \quad \forall g \in L_2 (Y).
$$
$$
\int_X f(\pi^\sharp g) \mathrm d \mu = \int_Y \left(\pi'_{\sharp} f\right) g \mathrm d \nu \quad \forall g \in L_1 (Y).
$$
Because $\nu$ is finite, we get $L_\infty(Y) \subset L_2(Y) \subset L_1(Y)$. So
- $\pi_{\sharp} f - \pi'_{\sharp} f \in L_2(Y)$.
- $$
\int_Y (\pi_{\sharp} f - \pi'_{\sharp} f) g \mathrm d \nu =0 \quad \forall g \in L_2 (Y).
$$
Pick $g:= \pi_{\sharp} f - \pi'_{\sharp} f$, we get
$$
\int_Y (\pi_{\sharp} f - \pi'_{\sharp} f)^2 \mathrm d \nu =0
$$
It follows that $\pi_{\sharp} f - \pi'_{\sharp} f=0$ $\nu$-a.e.
Let $\pmb 1 \in C(X)$ be the constant function. It suffices to prove that
$$
\int_X \pmb 1 \mathrm d \mu_y = 1 \quad \text{or equivalently} \quad \tilde{\pi}_{\sharp} \pmb 1 = 1 \quad \nu \text{-a.e.}
$$
By the duality, we have for all $g \in L_1(Y)$,
$$
\int_Y (\pi_\sharp \pmb 1) g \mathrm d \nu = \int_X \pmb 1(\pi^\sharp g) \mathrm d \mu = \int_X g \circ \pi \mathrm d \mu = \int_Y g \mathrm d \nu.
$$
It follows that for all $g \in L_1(Y)$
$$
\int_Y (\pi_\sharp \pmb 1 - 1) g \mathrm d \nu = 0.
$$
Because $\nu$ is finite, we get $L_\infty(Y) \subset L_1(Y)$. So $g := \pi_\sharp \pmb 1 - 1 \in L_1(Y)$. As such,
$$
\int_Y (\pi_\sharp \pmb 1 - 1)^2 \mathrm d \nu = 0.
$$
It follows that $\pi_\sharp \pmb 1 - 1 =0$ $\nu$-a.e. This completes the proof.
Best Answer
First of all there is no such thing as normal distribution on a compact metric space. Normal/Gaussian distribution id defined on Euclidean spaces or topological vector spaces but not on metric spaces. Your conclusion is true for Borel measures $nu $ on $X$ such that $\nu (U)>0$ for every non-empty open set $U$. A counter example to your statement is obtained by taking $\nu $ to be a delta measure at a point.