The answer is no. Endow both $\mathbb{R}^2$ and $\mathbb{R}$ with the usual Borel $\sigma$-algebra and let $\pi_1:\mathbb{R}^2\to\mathbb{R}$ be the projection onto the first coordinate. Now it is a well known fact that there exists Borel sets $B\subseteq\mathbb{R}^2$ such that $\pi_1(B)$ is not Borel. Now $B$ will be in general not open, but we can make it so. We just apply the following theorem (Lemma 4.58 in Alipranits & Border 2006):
Let $\mathcal{C}$ be a countable family of Borel subsets of a Polish space
$(X, \tau)$. Then there is a Polish topology $\tau'\supseteq\tau$ on $X$ with the same Borel $\sigma$-algebra for which each set in $\mathcal{C}$ is both open and closed.
In particular, we can make $B$ open. Then $B$ is open and $\pi_1$ is a continuous surjection, but $\pi_1(B)$ not Borel. Since one can take $B$ in the original space to be bounded, the example works even when $Y$ is compact.
It's false; $f(A)$ need not be Borel.
Let $C$ be the middle-thirds Cantor set. Define $\pi_1:C\times C\to C$ by $\pi_1(x,y)=x$.
True Fact There exists a Borel set $B\subset C\times C$ such that $\pi_1(B)$ is not Borel.
Hand Waving Well, that's well known with $\Bbb R$ in place of $C$. There simply cannot be any difference between $\Bbb R$ and $C$ in this regard, qed. Indeed it seems likely that the construction is going to be simpler for $C$. Look for instance at the sketch here. The sorts of codings needed there can certainly be done just as well with $C$ in place of $\Bbb R$. In fact when I think about natural ways to code well-orders, etc., as reals what I actually get are codings as elements of a Cantor set! For example, a natural way to code $R\subset\Bbb N\times \Bbb N$ as a real is $$R\mapsto\sum_{(n,m)\in R}3^{-2^n3^m},$$and that does map $R$ to an element of a certain Cantor set.
If someone who knows that stuff is inclined to confirm the true fact that would be fine.
Recall that $C$ is the set of all sums $$x=\sum_{j=1}^\infty x_j3^{-j}$$where each $x_j$ is $0$ or $2$. In defining various maps it will be convenient to write as though $$x=(x_1,x_2,\dots).$$Various inequalities below will be clear if you note that if $x,y\in C$ with $x\ne y$ then
$$|x-y|\sim 3^{-n},\quad n=\min\{j:x_j\ne y_j\}.$$
We begin. Let $h:C\to C\times C$ be the obvious bijection $$h(x)=((x_1,x_3,\dots),(x_2,x_4,\dots)).$$Note that $$|h(x)-h(y)|\le c|x-y|^{1/2}.$$Suppose $B\subset C\times C$ is a Borel set such that $\pi_1(B)$ is not Borel, and let $$A=h^{-1}(B).$$Let $$g=\pi_1\circ h.$$Now $g:C\to C$ is continuous, $A\subset C$ is Borel but $g(A)$ is not Borel.
But $g$ is not Lipschitz; we have just $|g(x)-g(y)|\le c|x-y|^{1/2}$. Here's the part that works for $C$ but not for $\Bbb R$: Define $s:C\to C$ by $$s(x)=(x_1,0,x_2,0,\dots).$$Then $s$ is injective, in particular $s$ is a homeomorphism from $C$ onto $s(C)$, and $s$ satisfies $$|s(x)-s(y)|\le c|x-y|^2.$$(Exercise: Show that if $\phi:\Bbb R\to\Bbb R$ satisifes $|\phi(x)-\phi(y)|\le c|x-y|^2$ then $\phi$ is constant. Figure out why that argument does not apply here.)
So we define $f:C\to C$ by $f=s\circ g$; then $f$ is Lipchitz but $f(A)$ is not Borel.
Finally, any Lipschitz $f:C\to C$ can easily be extended to a Lipschitz map $f:\Bbb R\to\Bbb R$. For example, define $f$ to be constant on $(-\infty,0]$, constant on $[1,\infty)$, and if $(a,b)$ is a bounded component of $\Bbb R\setminus C$ define $f$ to be linear on $[a,b]$.
Best Answer
Here, I tried to write the answer as explicit as possible. In fact, Noah Schweber already mentioned Lusin's example of analytic non-Borel set in the deleted answer. Also, Gio67 mentioned the Descriptive Set Theory notes. Thus, the answer is already there in Gio67's answer. I tried explaining the parts not covered by those answers.
Lemma
This is by continued frantion expansion of irrational numbers.
Definition
Theorem
Proof.
Let $\{x_n\}\in \mathbb{N}^{\mathbb{N}}$. Construct an increasing subsequence $\{n_k\}$ by using $\{x_{3k+1}\}$ as follows. \begin{align*} n_1&=x_1,\\ n_{k+1}&=n_k+x_{3k+1} \ \mathrm{for}\ k\geq 1. \end{align*} Then we place the numbers from $\{x_{3k+2}\}$ in the following way. \begin{align*} a_{n_1}&=x_2,\\ a_{n_{k+1}}&=a_{n_k}x_{3k+2}\ \mathrm{for}\ k\geq 1. \end{align*} Enumerate remaining numbers $\mathbb{N}-\{n_k\}=\{m_k\}$ in increasing order. Then use $\{x_{3k}\}$ to fill up these partial quotients. $$ a_{m_k}=x_{3k}\ \mathrm{for}\ k\geq 1. $$ Then define $f(\{x_n\})=[0;a_1,a_2,a_3,\ldots]$.
Now, we have an answer to the question.
Proof.
Consider the following composition. $$ g:I\rightarrow \mathbb{N}^{\mathbb{N}}\rightarrow E. $$ Then $G=\mathrm{Graph}(g)=\{(x,g(x))|x\in I\}$ forms a closed subset of a $G_{\delta}$ set $I\times [0,1]$. Then $G$ itself is a $G_{\delta}$ subset of $[0,1]^2$. Then the projection $\pi_2$ onto the second coordinate yields $\pi_2(G)=E$.
Let $\Phi:[0,1]\rightarrow [0,1]^2$ be the continuous space-filling curve. Then take the $G_{\delta}$ set $A=\Phi^{-1}(G)\subset [0,1]$. This gives $f(A)=E$ where $f=\pi_2\circ \Phi$.