$f:I\rightarrow X$ where $X$ is hausdorff show that $X$ is metrizable. This question comes from section 44 problem 4 of Munkres.
Let $X$ be a Hausdorff space. Let $I=[0,1]$. Show that if there is a continuous surjective map $f : I \rightarrow X$, then $X$ is compact, connected, weakly locally connected, and metrizable. [Hint: Show $f$ is a perfect map.]
I have already shown everything except that it is metrizable. I know that since $X$ is compact it is also paracompact, so I was going to use the Smirnov metrization theorem which states that if $X$ is paracompact and locally metrizable that $X$ is metrizable.
I am struggling with how to set up the neighborhood $U$ of $x\in X$, such that $U$ is metrizable. Any hints would be great in this direction.
Alternatively I thought about proving that $X$ was second-countable but I was also struggling with that. So any hints this direction would also be welcome.
 A: Sorry, my first answer is wrong!!!
Please allow me to correct the answer.
Situation.
$X$ is Hausdorff,
$(Y,d)$ is a compact metric space, and
$f:Y \to X$ is a continuous surjection.
Then $X$ is compact.
Definition.
Let $n\geq 1$ be an integer and $x,x'\in X$ a point.
Write
$$
\tilde{F}_n := \{ (y_1,y_2)\in Y\times Y | d(f^{-1}(f(y_1)),y_2) \geq 1/n\}.
$$
Then $\tilde{F}_n\subset Y\times Y$ is a closed symmetric subset.
Write $\Delta \subset X\times X$ for the diagonal.
Claim 1.

*

*$\tilde{F}_n \cap (f\times f)^{-1}(\Delta) = \emptyset$.

*$\bigcap_{n\geq 1}(Y\times Y \setminus \tilde{F}_n) = (f\times f)^{-1}(\Delta)$.

*$\bigcap_{n\geq 1}(X\times X \setminus (f\times f)(\tilde{F}_n)) = \Delta$.

Proof.
If $f(y_1) = f(y_2)$, then $y_2\in f^{-1}(f(y_1))$.
Hence $d(f^{-1}(f(y_1)),y_2) = 0$.
This implies the first equality.
The second equality follows from the compactness of $f^{-1}(f(y))$ (for any $y\in Y$).
The third equality follows from the second equality.
$\square$
Definition.
For any $\tilde{A}\subset X\times X$, we write $\tilde{A}^{\mathrm{op}} := \{(a,b)|(b,a)\in \tilde{A}\}$, and $\tilde{A}^{\mathrm{sym}} := \tilde{A}\cap \tilde{A}^{\mathrm{op}}$.
Write $\tilde{W}_n:= (X\times X \setminus (f\times f)(F_n))^{\mathrm{sym}}$.
Then $\tilde{W}_n$ is open and symmetric.
Write
$$
\mathcal{U}_n := \{U \ | \ U\text{ is open}, U\times U \subset \tilde{W}_n\}
$$
for the open covering of $X$.
Write
\begin{align*}
U_n(x) &:= \{ x'\in X | \forall y'\in f^{-1}(x'), \exists y\in f^{-1}(x), d(y,y') < 1/n\}, \\
W_n(x) &:= \{x'\in X | (x,x')\in \tilde{W}_n\}. 
\end{align*}
Claim 2.
For any $x\in X$ and any open subset $x\in U\subset X$,
there exists $n\geq 1$ such that $U_n(x)\subset U$.
Proof.
Take an integer $n \gg 1$ such that $d(f^{-1}(x),f^{-1}(X\setminus U)) > 1/n$.
Then $U_n(x) \subset U$.
$\square$
Claim 3.
$W_n(x) \subset U_n(x)$.
Proof.
Assume that $(x,x')\in W_n$.
Then $(f\times f)^{-1}(x,x') \cap F_n = \emptyset$.
Hence for any $y'\in f^{-1}(x')$, $d(f^{-1}(x),y') < 1/n$,
i.e., $\exists y\in f^{-1}(x), d(y,y') < 1/n$.
This implites that $x'\in U_n(x)$.
$\square$
Claim 4.
If $U\times U \subset W_n$, and $x\in U$, then $U\subset W_n(x)$.
Proof.
If $x'\in U$, then $(x,x')\in U\times U \subset W_n$.
Thus $x'\in W_n(x)$.
$\square$
Definition.
Since $X$ is compact (hence in particular, paracompact),
we can take a finite star (open) refinement
$\mathcal{V}_n$ of the open covering $\{U\cap V | U\in \mathcal{U}_n, V\in \mathcal{V}_{n-1}\}$, inductively.
Write
$$\mathcal{V}_n(x) := \bigcup \{V\in \mathcal{V}_n | x\in V\} \subset X.$$
Then it follows immediately that $\mathcal{V}_n(x) \subset \mathcal{V}_{n-1}(x)$.
Finally, we write $\mathcal{V} := \bigcup_{n\geq 1}\mathcal{V}_n$.
Then $\mathcal{V}$ is a countable set of open subsets of $X$.
Claim 5.
$\mathcal{V}$ is a basis of $X$.
In particular, $X$ is second countable.
Proof.
Let $x\in X$ be a point and $x\in U\subset X$ an open neighborhood of $x$.
By Claim 2 and Claim 3, there exists an integer $n\geq 1$ such that $W_n(x) \subset U$.
Since $\mathcal{V}_n$ is a star refinement of $\mathcal{U}_n$,
$\mathcal{V}_n(x) \times \mathcal{V}_n(x) \subset W_n$.
Hence, by Claim 4, $\mathcal{V}_n(x) \subset W_n(x) \subset U$.
This implies that $\exists V\in \mathcal{V}_n$ such that $x\in V \subset U$.
$\square$
Conclusion.
$X$ is metrizable.
Proof.
This follows directly from Uryshon metrization theorem. $\square$
A: The hint in Munkres recommends showing that $f$ is a perfect map then using Exercise 31.7 to show that since the interval is regular and second countable so is $f(I)$. A closed subset of $I$ is compact, so its image is compact in $X$.
We can prove $f$ perfect as follows: Since $X$ is Hausdorff, the image under $f$ of a closed set is not only compact but also closed. Thus, $f$ is a closed map. Further, since $X$ is Haussdorff, singleton sets are closed in $X$. Thus, by continuity of $f$ we see that $f^{-1}(\{x\})$ is a closed subset of $I$ for all $x\in X$. Since $I$ is compact, so are its closed subsets. Thus the preimages of single points are compact sets. $f$ has now been shown to be a closed map that is continuous and subjective in addition to the fibres of $f^{-1}$ being compact, so we conclude that $f$ is a perfect map.
From there it is not too hard to prove that the image under a perfect map of a first countable regular space is first countable and regular and thus metrizable by the Urysohn Metrization Theorem.
