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Exercise: Let $(X,tau)$ and $(Y,tau_1)$ be topological spaces and $f:(X,tau)to(Y,tau_1)$ a quotient mapping. If $(X,tau)$ is metrizable, is $(Y,tau_1)$ metrizable?

Attempted solution:

Lemma: A metrizable space is Hausdorff.

If I assume $(X,tau)$ to be a discrete space and $(Y,tau_1)$to be a topological space endowed with the cofinite topology. $(Y,tau_1)$ is not Hausdorff.

Now I need to check if $f$ defines the quotient topology $tau_1$. If ${y}in Y$ then it is closed so $f^{-1}({y})$ is closed then ${Y}$ is closed in the quotient topology such as the union of all singular points, then $tau_1$ is the cofinite topology.

Question:

Is this proof right? If not? Why not? What are the alternatives?

Thanks in advance!

It's somewhat unclear what you're trying to "prove". It seems like you want a counterexample where the quotient space is non-Hausdorff as that is an easy way to show it's not metrisable. Fine so far. But you do need to give a concrete example of that happening to fully solve the problem. So like @Jose in his answer you can take $X=mathbb{R}$ and the Vitali equivalence relation $xRy$ iff $x-y in mathbb{Q}$ and $f$ the quotient map from $X$ onto $Y=X/R$.

Another idea is to identify $mathbb{Z} subseteq mathbb{R}$ to a point and prove that the resulting quotient space is not first countable, also contradicting metrisability.

In either case, the proof must consist of an example and the proof that the example is as required. For both, proofs can be found on this site if you search for them.

The proof is not correct since at no point you orived that your mapping is a quotient mapping.

Take, for instance, $mathbb R$, endowed with the usual topology, and the quotient topology on $mathbb{R}/sim$, where$$xsim yiff x-yinmathbb{Q}.$$Then $mathbb{R}/sim$ is not Hausdorff and therefore not metrizable.