Let $X$ be connected and locally connected. Let $f:X\to [0,1]$ be bijective continuous function, then $X$ is homeomorphic to $[0,1]$ Let $X$ be connected and locally connected. Let $f:X\to [0,1]$ be continuous and bijective function, then $X$, $[0,1]$ are homeomorphic.
My attempt
Let $A\subset X$ be non-empty open set, let's see f(A) is open. 
we take $y\in f(A)$, then there exist $x\in A$ such that $f(x)=y$ ($f$ bijective).
By hypothesis $X$ is locally connected then exist $V\subset X$ open connected set such that $x\in V\subset A$, by continuity $f(V)$ is connected and $y\in f(V)\subset f(A)\subset  [0,1]$, thus $f(V)$ is an interval, then I think that is enough to prove that $f(V)$ is not of the form $[a,b]$, $[a,b)$, $(a,b]$ for $a>0$, $b<1$
 But I don't know how to prove it.
thanks in advance
 A: Let $I'$ the space with underlying set $I = [0,1]$ and topology $\mathfrak{T}_f = \{ f(U) \mid U \text{ open in } X \}$ (note that this is in fact a topology on $I$ because $f$ is a bijection). Then $f : X \to I'$ is a homeomorphism and the identity map $id : I' \to I$ is continuous which means that $\mathfrak{T}_f$ is finer than the standard topology $\mathfrak{T}$. We have to show that $\mathfrak{T}_f = \mathfrak{T}$.
Since $I'$ is locally connected, the connected open sets of $\mathfrak{T}_f$ form a base of $\mathfrak{T}_f$. Hence it suffices to show that each connected open set of $\mathfrak{T}_f$ belongs $\mathfrak{T}$.
So let $B$ be a connected open set of $\mathfrak{T}_f$. We have $B = id(B)$ connected in $I$, thus each $B$ is an interval. All intervals of the form $[0,1]$, $[0,a)$ with $0 < a \le 1$, $(a,b)$ with $0 \le a < b \le 1$ and $(a,1]$ with $0 \le a < 1$ belong to $\mathfrak{T}$. So let us consider whether other intervals can be open in $I'$.
1) $B = [a,b]$ with $0 \le a \le b \le 1$ and $\lvert b - a \rvert < 1$.
If it were open, then $V = ((-1,a) \cup (b, 2)) \cap [0,1]$ would be a nonempty open set in $\mathfrak{T }\subset \mathfrak{T}_f$ and $B, V$ would be a decomposition of $I'$ into nonempty disjoint open sets which is impossible. Hence $B$ cannot be open in $I'$.
2) $B = [a,b)$ with $0 < a < b \le 1$.
If it were open, then $B \cup (a,1] = [a,1]$ would be open in $I'$ which is impossible by 1).
3) $B = (a,b]$ with $0 \le a < b < 1$.
If it were open, then $B \cup [0,b) = [0,b]$ would be open in $I'$ which is impossible by 1).
