$X$ is $T_1$ implies $f(X)$ is $T_1$

Question

Let $X$ and $Y$ topological spaces, and $f:X \rightarrow Y$ a function such that

1. $Im(f) = Y$
2. If $A\subset X$ is closed, so $f(A)$ is closed in $Y$.

Let's suppose that $X$ is a $T_1$ space, is it possible to $Y$ be a $T_1$ space too?

(If $X$ is a $T_1$ space, then for $a,b \in X$, such that $a \not = b$, there are $U,V \subset X$ open subsets that

1. $a \in U$, $b\in V$
2. $b \not \in U$, $a \not \in V$)

Here is my attempt: If $y_1,y_2 \in Y$, $y_1 \not = y_2$, there are $x_1,x_2\in X$, $x_1 \not = x_2$, such that $f(x_i) = y_i, i = 1,2$. Since $X$ is $T_1$, there are open subsets $U,V \in X$, that $x_1 \in U, x_2 \in V$ and $x_1 \not\in V, x_2 \not \in U$. Since $U,V$ are open subsets, $X\setminus U, X \setminus V$ are closed subsets. Then, $f(X\setminus U),f(X\setminus V)$ are closed subsets in $Y$. The problem is that is not true that $$f(X\setminus U) = Y\setminus f(U).$$

Is there any hypothesis missing?

Answer 1

If $y \in Y$, we find $x \in X$ with $f(x)=y$.

As $X$ is $T_1$, $A=\{x\}$ is a closed subset of $X$, hence $f[A]=\{y\}$ is closed in $Y$. It follows that $Y$ is $T_1$ too.