There is no such continuum. See
Z. Waraszkiewicz, Sur un problème de M.H. Hahn, Fund. Math. 22 (1934) 180–205.
Waraszkiewicz constructed an uncountable family $W$ of continua in the plane called Waraszkiewicz spirals so that no continuum can be mapped continuously onto every continuum in $W$. Start with the space $X=S^1\cup [1,2]$ and consider the maps $f,g:X\to X$ where
- $f(x)=x$ if $x\in S^1$, $f(x)=\exp(4\pi ix)$ for $1\leq x\leq \frac{3}{2}$, and $f(x)=2(x-1)$ for $\frac{3}{2}\leq x\leq 2$
- $g(x)=x$ if $x\in S^1$, $g(x)=\exp(-4\pi ix)$ for $1\leq x\leq \frac{3}{2}$, and $g(x)=2(x-1)$ for $\frac{3}{2}\leq x\leq 2$
Now $W$ is the collection of all inverse limits of all inverse sequences $(X_i,h_i)$ where $X_i=X$ for all $i$ and $h_i\in\{f,g\}$.
For a short and readable description (where I found the construction) see:
W.T. Ingram, Concerning images of continua, Topology Proceedings 16 (1991) 89-93.
This non-existence results has been improved upon a few times, for instance in the following:
S.B. Nadler, The nonexistence of almost continuous surjections between certain continua, Topology and its Applications, 154 (2007) 1008-1014.
The answer to this question is negative. A suitable counterexample can be constructed as follows.
Let $Y=\mathbb R$ be the real line with the standard Euclidean topology.
Let $\mathbb Q$ be the subspace of rational numbers in $\mathbb R$. Write $\mathbb R\setminus \mathbb Q$ as the union $\bigcup_{q\in \mathbb Q}X_q$ of pairwise disjoint dense sets in $\mathbb R$. Let $$X=\mathbb Q\oplus\bigoplus_{q\in \mathbb Q}(\{q\}\cup X_q)$$be the topological sum of the spaces $\mathbb Q$ and $\{q\}\cup X_q$ for $q\in\mathbb Q$.
Let $\varphi:X\to Y$ be the natural projection. It is clear that the spaces $X,Y$ are separable, metrizable (and hence Tychonoff) and the function $\varphi:X\to Y$ is skeletal.
On the other hand, in the $\mathbb R$-quotient topology $\tau$ on $Y=\mathbb R$, the set $\mathbb Q$ is closed and nowhere dense in $(Y,\tau)$, witnessing that the function $\varphi:X\to(Y,\tau)$ is not skeletal.
To show that $\mathbb Q$ is closed in $(Y,\tau)$, choose any point $y\in\mathbb R\setminus \mathbb Q$. Find $q\in\mathbb Q$ such that $y\in X_q$. Consider the function $f_q:Y\to\mathbb R$ defined by
$f_q(x)=|x-q|$ if $x\in X_q$ and $f_q(x)=0$, otherwise.
Observe that the composition $f_q\circ \varphi:X\to\mathbb R$ is continuous, which implies that the set $X_q=\{x\in Y:f_q(x)>0\}$ is $\tau$-open, contain $y$, and does not intersect $\mathbb Q$.
This completes the proof of the closedness of $\mathbb Q$.
Assuming that the closed set $\mathbb Q$ is not nowhere dense in $(Y,\tau)$, we can find a nonempty open set $U\subseteq\mathbb Q$. Choose any point $q\in U$ and observe that the preimage $\varphi^{-1}(U)$ is an open set in $X$ containing the point $q\in \{q\}\cup X_q$. Since $\{q\}$ is nowhere dense in $X_q$, the set $\varphi^{-1}(U)$ has nonempty intersection with the set $X_q$ and then $\emptyset \ne \varphi[X_q\cap \varphi^{-1}(U)]\subseteq U\cap X_q\subseteq U\setminus\mathbb Q$, which contradicts the choice of $U\subseteq\mathbb Q$.
Best Answer
Call $A$ an HM-space if there is a continuous surjection $I\to A$ (where $I$ is the closed interval $[0,1]$). Note that if $A$ is an HM-space, then it is path-connected.
Theorem: If $X$ is path-connected, then the following are equivalent:
Proof: Clearly (2) and (3) are equivalent; let's prove (3) implies (1). If $f:(0,\infty) \to X$ is a continuous surjection, let $A_n = f([n-1,n])$. Then $A_n$ is evidently an HM-space and $X$ is the indicated countable union.
Now we show (1) implies (4). if $X = \bigcup_{n=1}^\infty A_n$ with each $A_n$ an HM-space, then we can choose surjective paths $\alpha_n: [2(n-1),2n-1]\to A_n$; write $x_n = \alpha_n(0)$ and $y_n= \alpha_n(1)$. Since $X$ is path-connected, we can find paths $\beta_n:[2n-1,2n]\to X$ from $y_n$ to $x_{n+1}$. Concatenating these paths gives the desired surjection $0,\infty) \to X$.
Finally, since there is a continuous surjection $\mathbb{R}\to [0,\infty )$, (4) implies (2).$\quad \square$
Just musing: there is a partial order on spaces defined by $X< Y$ if there is a continuous surjection $Y \to X$. Hahn-Mazurkowicz classifies $\{ X \mid X < I\}$. The equivalence of (2), (3) and (4) shows that half-open intervals are "equivalent" to open ones (i.e., both $(0,1)< [0,1)$ and $[0,1) < (0,1)$); and closed ones are different, being compact. Finally, note that a Peano space (i.e., HM-space) $X$ is "equivalent" in this sense to $I$ if and only if it is the domain of a nonconstant continuous function $f:X\to \mathbb{R}$.