Let $(X,d)$ be a metric space, and let $f$ be a real valued function defined on $X$. Let $f_n$ be a sequence of real valued functions defined on $X$.
Fix $u \in X$
Assume that for every $\delta > 0, \limsup_{j\to\infty} \inf\{f_j(v): d(u,v) <\delta \} \leq f(u)$
Then, the claim is that there exists a sequence $\{u_j\}$ which converges to $u$ in $X$ such that $$\limsup_{j\to\infty} f_j(u_j) \leq f(u) $$
To prove the claim, here is wht I tried:
Let $u_n$ such that
$$f_n(u_n) < \inf \{f_n(v): d(u,v) < \frac{1}{n} \} + \frac{1}{n} $$ Then, I took the limsup of both sides, but then my "delta" $\frac{1}{n}$ shrunk to zero, so I cannot conclude what I want.
How should I proceed?
The problem becomes more transparent if it is recast in terms of the epigraph of $f_j$, which is the set $$E_j = \{(v,y): y\ge f_j(v) \}\subset X\times \mathbb{R}$$ The given condition says: for every $\epsilon>0$ and every $\delta>0$, for all large enough $j$ there is a point $(v,y)\in E_j$ such that $d(u,v)<\delta$ and $y = f(u)+\epsilon$. (Indeed, this is saying that $f_j(v)\le f(u)+\epsilon$.)
We can simplify this by using $\epsilon=\delta$, and then reword the assumption concisely as $$\operatorname{dist}((u,f(u)), E_j)\to 0 \tag{1}$$ (This refers to a metric on the product $X\times \mathbb{R}$; for example the metric $\rho ((v,y),(w,z)) = d(v,w) + |y-z|$ will do.)
Now the solution is immediate: pick a point $(v_j,y_j)\in E_j$ such that $$\rho( (u,f(u)), (v_j,y_j)) < \operatorname{dist}((u,f(u)),E_j) +\frac1j $$ Then $\rho( (u,f(u)), (v_j,y_j))\to 0$, which implies both $v_j\to u$ and $\limsup f(v_j) \le f(u)$.