That's not quite the same thing.
What I said is:
$y \in Ha \cap Hb \implies Ha = Hb$
What you just said (in your last post) is:
$y \in Hb \implies y \in Ha$
which is NOT equivalent.
Your last statement is equivalent to saying $Hb \subseteq Ha$, which does not, in and of itself, force equality of the two cosets (we must also have $Ha \subseteq Hb$).
But what I said is an even STRONGER statement:
If two cosets have ANY element in common, they are the SAME coset.
*********
A typical proof runs something like this:
Suppose $y \in Ha \cap Hb$. This means that $y = ha$ for some $h \in H$, and that $y = h'b$ for some (typically different) element $h' \in H$.
From:
$ha = h'b$
we see that:
$b = h'^{-1}ha$
so that for any element $h''b \in Hb$, we have:
$h''b = (h''h'^{-1}h)a \in Ha$ since $H$ is a subgroup, and thus closed under multiplication and inversion.
This shows that $Hb \subseteq Ha$.
The proof that $Ha \subseteq Hb$ is similar, using the fact that $a = h^{-1}h'b$.
************
Another way to state these facts are:
Two (right) cosets of $H$ are either the same, or disjoint.
To see, this, note that what I have shown above is that if two cosets are not disjoint (having the common element $y$), they are equal.
