Isomorphism between group of homeomorphisms where $X nsim Y$

This is question 7N #2 from Willard’s General Topology, on p. 51.

For any topological space $X$, let $H(X)$ denote the group of
homeomorphisms of $X$ onto itself, with composition as the group
operation. Let $X = [0,1]$ and $Y = (0,1)$ and define $phi(h) = hrestriction_Y$, for each $h in H(X)$. Then $phi$ is an isomorphism
of $H(X)$ with $H(Y)$ but there is no homeomorphism of $X$ onto $Y$.

I have already seen that $[0,1]$ is not homeomorphic to $(0,1)$ (i.e. $X nsim Y$), so I am just trying to show that $phi: H(X) to H(Y)$ is an isomorphism. I am hoping that someone can double check my proof for showing that $phi$ preserves the composition operation and that $phi$ is injective and hoping that someone can help me show that $phi$ is also surjective, which I am struggling with.

First, let $h_1, h_2 in H(X)$. Then:

$phi(h_1 circ h_2) = (h_1 circ h_2)restriction_Y = (h_1restriction_Y) circ (h_2restriction_Y) = phi(h_1) circ phi(h_2)$.

I’m not sure if the move from the first equality to the second is obvious. It seems like it should be but I’m not sure if I need to justify it more. If this argument works, it shows that $phi$ preserves the group operation.

Second, suppose $h_1, h_2 in H(X)$ and $phi(h_1) = phi(h_2)$. This implies that $h_1restriction_Y = h_2restriction_Y$, i.e. $h_1(x) = h_2(x) (forall x in (0,1) = Y)$. But since $h_1, h_2$ are continuous on $X = [0,1]$, then $h_1(x) to h_1(0)$ and $h_2(x) to h_2(0)$ as $x to 0$. Since $h_1(x) = h_2(x) forall x in (0,1)$, then they must converge to the same point, by uniqueness of limits. So $h_1(0) = h_2(0)$. A similar argument holds when taking $x to 1$, so we can conclude that $h_1(x) = h_2(x) forall x in [0,1]$, i.e. $h_1 = h_2$ on $X$, so $phi$ is injective.

It remains to show that $phi$ is surjective. This is where I’m less sure.

My idea is to take some $g in H(Y)$ and a sequence like $(x_n) = (frac{1}{n+1}), n in mathbb{N}$. Clearly $x_n to 0$. Since $x_n in Y$ for each $n in mathbb{N}$ then we can define the sequence $(g_n) = (g(frac{1}{n+1})), n in mathbb{N}$. Since $g$ is continuous, we have that $(g_n) to x_0$ for some $x_0 in X$ (I think?). We can do something similar starting with a sequence in $Y$ that converges to $1$. So if we want to extend $g$ to a homeomorphism $h$ in $H(X)$ we would take $h(x) = g(x)$ for $x in Y$, $h(0) = x_0$, and $h(1) = x_1$. Does this argument make sense? Intuitively, it seems like it should be correct but I’m worried I’m not being rigorous enough.