There is a famous theorem that for a compact topological space $X$, continuous bijection $f:X\to Y $ where $Y$ is Hausdorff is homeomorphic.
I want to know if the converse is true, i.e.
Let topological sapce $X$ satisfy the following property:
For any continuous bijection $f:X\to Y $ where $Y$ is Hausdorff is homeomorphic.
Then, can we say $X$ is compact?
For example, if $X=(0,1]$ then let $Y$ be a circle and continuous bojection $f:X\to Y$ be connecting $1\in X$ to $0\notin X$, so we fail to construct counterexample. If $X=(0,1)$ then connect $0\notin X$ to $\frac{1}{2}\in X$.
If $f:X\to Y$ is continuous then we have a new topology on $X$ generated by subsets of the form $f^{-1}(U)$ over open subsets $U\subseteq Y$. This topology is coarser than that of $X$, and $f$ (seen as a map onto its image) is a quotient map under this topology.
So if $f$ is bijective and a quotient map than it is a homeomorphism. Under the new topology of course. So your question can be restated as:
Here "minimal" means coarsest among Hausdorff topologies. Note that a compact Hausdorff space indeed is minimal.
The answer to that question is "no". In other words there is a non-compact Hausdorff space $X$ such that every continuous bijection starting at $X$ onto a Hausdorff space is a homeomorphism. A concrete example can be found in the famous Counterexamples in Topology book, part II, counterexample 100. Also see this: https://mathoverflow.net/questions/36085/minimal-hausdorff