Is this a sufficient condition for two spaces to be homeomorphic; proof or counter example please.

No; see the counterexamples on this MathOverflow thread.

Gerhard Paseman gives this counterexample:

Let $X = Y = \mathbb{Z} \times \{0,1\}$. We declare that the following subsets of $X$ are open for each $n>0$: $$\{(-n,0)\},\quad \{(-n,1)\},\quad \{(0,0)\},\quad \{(0,0),(0,1)\},\quad \{(n,0),(n,1)\}$$ This is a basis for a topology on $X$.

We declare that the following subsets of $Y$ are open for each $n>0$: $$\{(-n,0)\},\quad \{(-n,1)\},\quad \{(0,0),(0,1)\},\quad \{(n,0),(n,1) \}$$ This is a basis for a topology on $Y$.

Define $f:X\to Y$ and $g:Y\to X$ by $f((n,i))=(n,i)$ and $g((n,i))=(n+1,i)$. Then $f$ and $g$ are continuous bijections, but $X$ and $Y$ are not homeomorphic.

Scott Carnahan gives this counterexample:

Here's a continuum analogue of Gerhard Paseman's answer: Let $X$ and $Y$ be topological spaces whose underlying sets are $\mathbb{R}$. As topological spaces, $X$ is the disjoint union of the open interval $(0,\infty)$ with a discrete space whose points are nonpositive reals, while $Y$ is the disjoint union of $(-1,0)$, $(1,\infty)$, and a discrete space whose points form the complement of those intervals. Translation by adding one is a continuous bijection from $X$ to $Y$, and also a continuous bijection from $Y$ to $X$, but the two spaces are not homeomorphic.