Continuous image of a locally path connceted space is locally path connected? We already know that continuous image of path-connected spaces are path-connceted. I was thinking if the same result is true when we replace path - connected with locally path - connceted.
I think I have proved this for a continuous and open map.

Is the openness requirement necessary or continuity is enough (and simply I was not able to prove the statement in that case)?

More precisely, this is what I proved: suppose $f:X \to Y$ is a continuous and open map. Let $a=f(x) \in f(X)$ and let $J$ be a neighborhood bases for $x$ of path- connected spaces. Then $J'=\{f(A): A \in J\}$ is a neighborhood bases for $a$ of path- connected spaces. The openness of $f$ helped me to say that $A$ neighborhood of $x$ implies $f(A)$ neighborhood of $a$. The continuity of $f$ implies that $f(A)$ is path - connceted whenever $A$ is path - connected. Finally, if $U$ is a neighborhood of $a$ then $f^{-1}(U)$ is a neighborhood of $x$ and then there is $A\in J: A\subseteq f^{-1}(U)$, then $a \in f(A) \subseteq U$.
 A: I think openness is necessary. For instance take a totally disconnected space $S$ that is not discrete, say it has no open points. For the sake of concreteness, you may take $S= 2^A$ for some infinite set $A$ with the discrete topology, and $S$ has the product topology. 
Let $I$ denote the standard interval $[0,1]$.
Let $Y= S\times I /( (s,1)\sim (t,1), s,t\in S)$ with the quotient topology of the product topology and call $\infty$ the image of the $(t,1)$'s in the quotient. 
Quite clearly $Y$ is path-connected as every point is connected to $\infty$. 
It's also easy to check that we may identify $S$ with the image of $S\times \{0\}$ in $Y$. 
Now your space $Y$ is clearly not locally path-connected : take $s\in S$, and an open neighbourhood $U$ of $s\in S$. Then it contains some $V\times [0,\epsilon)$ for some neighbourhood $V$ of $s\in S$. 
But $S$ has no open point, so there  is $t\in V\setminus\{s\}$, and any path from $s$ to $t$ in $V\times [0,\epsilon)$ has a continuous projection on $S\times \{0\}$, which is absurd because $S$ is totally disconnected. 
So our space $Y$ is a good candidate, because it is path-connected but not locally path-connected. 
Now let $X$ have the same underlying set as $Y$, but this time we put the discrete topology on $S$ (and then the product topology on $S\times I$, and then the quotient one on $X$). Then $X$ is path-connected (same reason, any point is connected to $\infty$), and locally path-connected : if you take $(s,t)\in X$, then $\{s\}\times ]t-\epsilon, t+\epsilon[$ is an open path-connected neighbourhood of $(s,t)$ and those are a basis of neighbourhoods at $(s,t)$, if $t\in (0,1)$, and I'll let you see how this works for $t=0, 1$ (for $t=0$ it's the same thing, and for $1$ it's a tiny bit more subtle, but really not that much)
But now there is a continuous bijection $X\to Y$ : it's simply the identity on underlying sets ! Its continuity comes from the fact that any map from a discrete space is continuous, and from various universal properties (of the product, and of the quotient topology). 
Clearly it's not open, but this shows that the new statement is false (though your proof in the open case is fine)
