"Unique Limit" involving continuous functions and Hausdorff space I want to prove the following:

Suppose $(X,\tau)$ is a topological space. $X$ has the property that whenever $f:(Y,\sigma)\to (X,\tau)$ is a continuous function defined on a topological space $Y$, and $b$ a limit point of $Y$, and $f(x)\to l$ as $x\to b$, then $l$ is unique. Show that $X$ is Hausdorff. Show also that the converse is true.

I find this rather difficult because it has a universal quantifier on $Y$ and $f$, which are not elements of any topological spaces. I end up partially proving the converse only.
Partial Proof of converse. Suppose $f(x)\to l,f(x)\to m, l\neq m$, but $X$ is Hausdorff. Then there exists open neighborhoods $U,V$ of $l,m$ such that $U\cap V=\emptyset$. Since $f$ is continuous, the preimages $f^{-1}(U), f^{-1}(V)$ are open sets. I guess, but cannot prove that $f^{-1}(U), f^{-1}(V)$ are disjoint neighborhoods of $b$, which is not possible since $b$ is a limit point.
But that logic doesn't seem to work for the other direction of the argument.
So how can I prove the whole statement?
EDIT I have now proved the converse part. How to prove the forwards direction?
 A: So formally, $X(,\tau)$ has the property "unique function limits":

UFL: For all spaces $(Y, \sigma)$ and all continuous $f:(Y, \sigma) \to (Y,\tau)$, and all $b \in Y$ that are limit points of $Y$ and all $l, l' \in X$  we have that $$(\lim_{x \to b} f(x) = l) \land (\lim_{x \to b} f(x)=l') \implies l=l'$$

And you want to show $$\textbf{UFL} \iff X \text{ Hausdorff }$$
You already started on the reverse: supposing the Hausdorff property and the setting as needed for UFL, so we have $Y$, $b$ and $\lim_{x \to b} f(x) = l$ and $\lim_{x \to b} f(x)=l'$ for some $l,l' \in Y$; Assume for a contradiction that $l \neq l'$ and take disjoint neighbourhoods $U$ of $l$ and $V$ of $l'$ such that $U \cap V = \emptyset$. Applying the limit to $l$ property we get a neighbourhood $W_1$ of $b$ such that $$f[W_1 \setminus \{b\}] \subseteq U$$ and from the limit with value $l'$ we get a neighbourhood $W_2$ of $b$ such that 
$$f[W_2 \setminus \{b\}] \subseteq V$$ 
which implies that $W_1 \cap W_2$ is an open neighbourhood of $b$ such that 
$$f[(W_1 \cap W_2) \setminus \{b\}] \subseteq U \cap V=\emptyset \text{ so }W_1 \cap W_2=\{b\} $$ so that $b \notin Y'$, a contradiction to the setup for UFL, so indeed $l=l'$ and so function limits are unique if they exist. Note that continuity of $f$ is not really needed, just the limits must exist.
For the forward direction: suppose UFL holds and we want to show $X$ is Hausdorff. So suppose not: then there are two distinct points $p \neq q $ of $X$ such that 
$$\forall U \in \mathcal{O}(p): \forall V \in \mathcal{O}(q): U \cap V \neq \emptyset\tag{1}$$ where $\mathcal{O}(x)$ is the set of open neighbourhoods of $x$ for $x \in X$. Now try to define a "custom made" $Y$ and $f$ to contradict UFL, using the structure of neighbourhoods intersecting in $X$ (filters are an option, or nets/directed sets, if you know them), so that $\lim_{x \to b} f(x)=p$ and also $\lim_{x \to b} f(x)=q$ for some limit point  $b$ in that $Y$. This would be the required contradiction and $X$ must be Hausdorff after all.
A: You've almost got the converse, you need only observe if $f^{-1}(U) \cap f^{-1}(V) \neq \emptyset$, then there exists $y \in (f^{-1}(U) \cap f^{-1}(V))\subset Y$. What is $f(y)$? It must reside in $U$ and $V$, but that isn't possible by construction of $U$ and $V$, as we constructed them to be disjoint, hence $U \cap V = \emptyset$. (We are using the definition of function to conclude that $f^{-1}(U) \cap f^{-1}(V) = \emptyset$, that the inverse image of disjoint sets is disjoint is a property of functions themselves; try proving it yourself, it's not actually that hard).
Edit:
I'm having trouble with the forwards direction, but one thing that's really useful is that it is enough to show that $(X, \tau )$ has unique limits (when they exist) and that it's first countable. Those two properties together imply Hausdorff. Showing uniqueness of limits, when they exist, is easy. Let $(Y,\sigma)=(X,\tau)$ and $f$ be the identity map on $X$. Now if we have a limit point $b$ and sequence $x_k\rightarrow b$, then we have $x_k=f(x_k)\rightarrow l=b$. f is trivially continuous in this case, and the hypothesis implies that the limit must be unique. (I'm abusing the fact that the hypothesis is universally quantified, to pick on a helpful example of a $Y$ and $f$ that is continuous). This works, because the map is trivial, any sequence in $Y$ that has a limit is also a sequence in $X$ (namely itself), that also has a limit (which is the same point), and the hypothesis says that with these sequences and continuous f, we can conclude that the limit is unique. This leaves only establishing that $X$ is first countable.
